💡Okay, so I have this expression: (x^8 - 81). I need to factor it completely. Hmm, let's see. I remember that when factoring expressions, especially polynomials, it's helpful to look for patterns like differences of squares or cubes. First, I notice that both (x^8) and 81 are perfect squares. Wait, (x^8) is the same as ((x^4)^2), right? And 81 is (9^2). So, this looks like a difference of squares because it's in the form (a^2 - b^2). I remember that the difference of squares can be factored as ((a - b)(a + b)). So, applying that here, I can write:[x^8 - 81 = (x^4)^2 - 9^2 = (x^4 - 9)(x^4 + 9)]Alright, that's the first step. Now, I have two factors: (x^4 - 9) and (x^4 + 9). Let me look at each of them to see if they can be factored further.Starting with (x^4 - 9), I notice that this is also a difference of squares because (x^4) is ((x^2)^2) and 9 is (3^2). So, applying the difference of squares formula again:[x^4 - 9 = (x^2)^2 - 3^2 = (x^2 - 3)(x^2 + 3)]Great, now I've broken down (x^4 - 9) into two more factors: (x^2 - 3) and (x^2 + 3). Let me check if these can be factored further.Looking at (x^2 - 3), this is another difference of squares because (x^2) is (x^2) and 3 is (sqrt{3}^2). So, I can factor it as:[x^2 - 3 = (x - sqrt{3})(x + sqrt{3})]But wait, if I'm factoring over the real numbers, I can stop here because (sqrt{3}) is a real number. However, if I were factoring over the complex numbers, I might need to go further, but I think the problem expects real factors.Now, looking at (x^2 + 3), this is a sum of squares. I remember that sums of squares don't factor nicely over the real numbers because there's no real number that squares to a negative number. So, (x^2 + 3) is prime over the reals and can't be factored further.Going back to the other factor, (x^4 + 9). This is a sum of squares as well because (x^4) is ((x^2)^2) and 9 is (3^2). Similar to the previous case, sums of squares don't factor over the real numbers. But wait, maybe I can factor it over the complex numbers? Let me think.If I consider complex numbers, I can factor (x^4 + 9) further. I recall that (x^4 + 9) can be written as (x^4 + 9 = (x^2)^2 + (3)^2), which is a sum of squares. To factor this, I can use the identity:[a^2 + b^2 = (a + bi)(a - bi)]So, applying that here:[x^4 + 9 = (x^2 + 3i)(x^2 - 3i)]But these are still quadratic factors. If I want to factor them completely over the complex numbers, I can factor each quadratic into linear terms. Let's try that.Starting with (x^2 + 3i), I can write it as:[x^2 + 3i = (x + sqrt{3i})(x - sqrt{3i})]But wait, (sqrt{3i}) is a complex number. To find its exact form, I might need to express it in terms of real and imaginary parts. Similarly, for (x^2 - 3i), it would be:[x^2 - 3i = (x + sqrt{-3i})(x - sqrt{-3i})]This is getting a bit complicated, and I'm not sure if the problem expects complex factors. Maybe I should stick to real factors unless specified otherwise.So, to recap, over the real numbers, the factorization of (x^8 - 81) is:[(x^4 + 9)(x^2 + 3)(x^2 - 3)]And over the complex numbers, it would be:[(x^4 + 9)(x^2 + 3)(x^2 - 3) = (x^2 + 3i)(x^2 - 3i)(x^2 + 3)(x^2 - 3)]But if I factor each quadratic further into linear terms over the complex numbers, it would be:[(x + sqrt{3}i)(x - sqrt{3}i)(x + sqrt{3})(x - sqrt{3})(x + sqrt{3}i)(x - sqrt{3}i)]Wait, that doesn't seem right. I think I made a mistake there. Let me correct that.Actually, (x^4 + 9) can be factored into two quadratics over the real numbers using the sum of squares formula with complex numbers, but to factor it completely into linear terms, I need to find the roots of (x^4 + 9 = 0). Solving (x^4 = -9), we can write (-9) in polar form as (9(cos pi + i sin pi)). Then, using De Moivre's theorem, the fourth roots would be:[x = sqrt[4]{9} left( cosleft( frac{pi + 2kpi}{4} right) + i sinleft( frac{pi + 2kpi}{4} right) right) quad text{for } k = 0, 1, 2, 3]Calculating these, we get four complex roots. Therefore, (x^4 + 9) factors into four linear terms over the complex numbers:[(x - sqrt[4]{9} e^{ipi/4})(x - sqrt[4]{9} e^{i3pi/4})(x - sqrt[4]{9} e^{i5pi/4})(x - sqrt[4]{9} e^{i7pi/4})]But this is getting quite involved, and I'm not sure if the problem requires this level of factoring. It might be sufficient to factor it into real quadratic factors.So, perhaps the complete factorization over the real numbers is:[(x^4 + 9)(x^2 + 3)(x^2 - 3)]And over the complex numbers, it would include the linear factors as well.Wait, but I think I can factor (x^4 + 9) further over the real numbers by recognizing it as a sum of squares, which can be factored into two real quadratics. The formula for factoring (a^4 + b^4) is ((a^2 + sqrt{2}ab + b^2)(a^2 - sqrt{2}ab + b^2)), but I'm not sure if that applies here.Alternatively, I recall that (x^4 + 9) can be factored as ((x^2 + sqrt{3}x + 3)(x^2 - sqrt{3}x + 3)). Let me check if that's correct.Multiplying these two quadratics:[(x^2 + sqrt{3}x + 3)(x^2 - sqrt{3}x + 3) = x^4 - (sqrt{3}x)^2 + 3x^2 + 3x^2 - 3x^2 + 9]Wait, that doesn't seem right. Let me do it step by step.First, multiply (x^2) by each term in the second quadratic:[x^2 cdot x^2 = x^4][x^2 cdot (-sqrt{3}x) = -sqrt{3}x^3][x^2 cdot 3 = 3x^2]Next, multiply (sqrt{3}x) by each term:[sqrt{3}x cdot x^2 = sqrt{3}x^3][sqrt{3}x cdot (-sqrt{3}x) = -3x^2][sqrt{3}x cdot 3 = 3sqrt{3}x]Finally, multiply 3 by each term:[3 cdot x^2 = 3x^2][3 cdot (-sqrt{3}x) = -3sqrt{3}x][3 cdot 3 = 9]Now, combine all these terms:[x^4 - sqrt{3}x^3 + 3x^2 + sqrt{3}x^3 - 3x^2 + 3sqrt{3}x + 3x^2 - 3sqrt{3}x + 9]Simplify by combining like terms:- The (x^4) term remains: (x^4)- The (-sqrt{3}x^3) and (sqrt{3}x^3) cancel out: (0)- The (3x^2 - 3x^2 + 3x^2) simplifies to (3x^2)- The (3sqrt{3}x - 3sqrt{3}x) cancel out: (0)- The constant term is (9)So, combining these, we get:[x^4 + 3x^2 + 9]Wait, that's not equal to (x^4 + 9). There's an extra (3x^2) term. So, my initial assumption about the factorization was incorrect. Maybe I need a different approach.Perhaps I should use the sum of squares formula with complex numbers but express it in terms of real quadratics. I recall that (x^4 + 9) can be factored as ((x^2 + sqrt{3}x + 3)(x^2 - sqrt{3}x + 3)). Let me verify this.Multiplying ((x^2 + sqrt{3}x + 3)(x^2 - sqrt{3}x + 3)):First, multiply (x^2) by each term in the second quadratic:[x^2 cdot x^2 = x^4][x^2 cdot (-sqrt{3}x) = -sqrt{3}x^3][x^2 cdot 3 = 3x^2]Next, multiply (sqrt{3}x) by each term:[sqrt{3}x cdot x^2 = sqrt{3}x^3][sqrt{3}x cdot (-sqrt{3}x) = -3x^2][sqrt{3}x cdot 3 = 3sqrt{3}x]Finally, multiply 3 by each term:[3 cdot x^2 = 3x^2][3 cdot (-sqrt{3}x) = -3sqrt{3}x][3 cdot 3 = 9]Now, combine all these terms:[x^4 - sqrt{3}x^3 + 3x^2 + sqrt{3}x^3 - 3x^2 + 3sqrt{3}x + 3x^2 - 3sqrt{3}x + 9]Simplify by combining like terms:- (x^4)- (-sqrt{3}x^3 + sqrt{3}x^3 = 0)- (3x^2 - 3x^2 + 3x^2 = 3x^2)- (3sqrt{3}x - 3sqrt{3}x = 0)- (9)So, the result is:[x^4 + 3x^2 + 9]Again, this is not equal to (x^4 + 9). There's an extra (3x^2) term. Hmm, so my factorization is incorrect. Maybe I need to adjust the coefficients.Alternatively, perhaps I should use the identity for sum of squares in a different way. I remember that (a^4 + b^4) can be factored as ((a^2 + sqrt{2}ab + b^2)(a^2 - sqrt{2}ab + b^2)). Let's try applying that here.Let (a = x) and (b = sqrt{3}), then:[x^4 + (sqrt{3})^4 = x^4 + 9]So, using the identity:[x^4 + 9 = (x^2 + sqrt{2} cdot x cdot sqrt{3} + (sqrt{3})^2)(x^2 - sqrt{2} cdot x cdot sqrt{3} + (sqrt{3})^2)]Simplify:[= (x^2 + sqrt{6}x + 3)(x^2 - sqrt{6}x + 3)]Let me check this multiplication:[(x^2 + sqrt{6}x + 3)(x^2 - sqrt{6}x + 3)]Multiply term by term:- (x^2 cdot x^2 = x^4)- (x^2 cdot (-sqrt{6}x) = -sqrt{6}x^3)- (x^2 cdot 3 = 3x^2)- (sqrt{6}x cdot x^2 = sqrt{6}x^3)- (sqrt{6}x cdot (-sqrt{6}x) = -6x^2)- (sqrt{6}x cdot 3 = 3sqrt{6}x)- (3 cdot x^2 = 3x^2)- (3 cdot (-sqrt{6}x) = -3sqrt{6}x)- (3 cdot 3 = 9)Combine like terms:- (x^4)- (-sqrt{6}x^3 + sqrt{6}x^3 = 0)- (3x^2 - 6x^2 + 3x^2 = 0)- (3sqrt{6}x - 3sqrt{6}x = 0)- (9)So, the result is:[x^4 + 9]Perfect! So, the correct factorization of (x^4 + 9) over the real numbers is:[(x^2 + sqrt{6}x + 3)(x^2 - sqrt{6}x + 3)]Therefore, going back to the original expression:[x^8 - 81 = (x^4 + 9)(x^2 + 3)(x^2 - 3)]And now, factoring (x^4 + 9) further:[= (x^2 + sqrt{6}x + 3)(x^2 - sqrt{6}x + 3)(x^2 + 3)(x^2 - 3)]So, over the real numbers, this is the complete factorization. Each quadratic factor is irreducible over the reals.If I were to factor it completely over the complex numbers, I would need to factor each quadratic into linear terms. Let's do that for completeness.Starting with (x^2 - 3), we already factored it as:[(x - sqrt{3})(x + sqrt{3})]Now, for (x^2 + 3), the roots are complex:[x = pm sqrt{-3} = pm isqrt{3}]So, it factors as:[(x - isqrt{3})(x + isqrt{3})]Next, let's factor (x^2 + sqrt{6}x + 3). To find its roots, use the quadratic formula:[x = frac{-sqrt{6} pm sqrt{(sqrt{6})^2 - 4 cdot 1 cdot 3}}{2 cdot 1} = frac{-sqrt{6} pm sqrt{6 - 12}}{2} = frac{-sqrt{6} pm sqrt{-6}}{2} = frac{-sqrt{6} pm isqrt{6}}{2}]So, the factors are:[left(x - frac{-sqrt{6} + isqrt{6}}{2}right)left(x - frac{-sqrt{6} - isqrt{6}}{2}right) = left(x + frac{sqrt{6}}{2} - frac{isqrt{6}}{2}right)left(x + frac{sqrt{6}}{2} + frac{isqrt{6}}{2}right)]Similarly, for (x^2 - sqrt{6}x + 3), the roots are:[x = frac{sqrt{6} pm sqrt{6 - 12}}{2} = frac{sqrt{6} pm isqrt{6}}{2}]So, the factors are:[left(x - frac{sqrt{6}}{2} + frac{isqrt{6}}{2}right)left(x - frac{sqrt{6}}{2} - frac{isqrt{6}}{2}right)]Putting it all together, the complete factorization over the complex numbers is:[(x - sqrt{3})(x + sqrt{3})(x - isqrt{3})(x + isqrt{3})left(x + frac{sqrt{6}}{2} - frac{isqrt{6}}{2}right)left(x + frac{sqrt{6}}{2} + frac{isqrt{6}}{2}right)left(x - frac{sqrt{6}}{2} + frac{isqrt{6}}{2}right)left(x - frac{sqrt{6}}{2} - frac{isqrt{6}}{2}right)]That's quite a mouthful! But it shows all the linear factors over the complex numbers.However, since the original problem didn't specify, I think it's safe to assume they want the factorization over the real numbers. So, the complete factorization over the reals is:[(x^2 + sqrt{6}x + 3)(x^2 - sqrt{6}x + 3)(x^2 + 3)(x^2 - 3)]But wait, earlier I had (x^4 + 9) factored into those two quadratics, and then the other two quadratics from (x^4 - 9). So, combining all, the complete factorization is as above.Just to make sure, let me count the degrees. The original expression is degree 8. Each quadratic is degree 2, and there are four of them, so 2*4=8. That matches. Also, if I multiply them all out, I should get back to (x^8 - 81). Let me do a quick check.Multiplying (x^2 + sqrt{6}x + 3) and (x^2 - sqrt{6}x + 3) gives (x^4 + 9), as we saw earlier. Then, multiplying (x^2 + 3) and (x^2 - 3) gives (x^4 - 9). So, multiplying (x^4 + 9) and (x^4 - 9) gives (x^8 - 81), which is the original expression. So, that checks out.Therefore, the complete factorization over the real numbers is:[(x^2 + sqrt{6}x + 3)(x^2 - sqrt{6}x + 3)(x^2 + 3)(x^2 - 3)]And over the complex numbers, it's the product of all eight linear factors as I listed earlier.But since the problem doesn't specify, I'll present the real factorization as the answer.

💡Okay, so I need to find the greatest common divisor (GCD) of 7! and 10! divided by 4!. Hmm, let me think about how to approach this. I remember that the GCD of two numbers is the largest number that divides both of them without leaving a remainder. So, I need to figure out what divides both 7! and 10!/4!.First, maybe I should compute what 7! and 10!/4! actually are. Let me start by calculating 7!. I know that 7! means 7 factorial, which is 7 × 6 × 5 × 4 × 3 × 2 × 1. Let me compute that:7! = 7 × 6 × 5 × 4 × 3 × 2 × 1= 7 × 6 = 4242 × 5 = 210210 × 4 = 840840 × 3 = 25202520 × 2 = 50405040 × 1 = 5040So, 7! is 5040.Now, let me compute 10! and then divide it by 4!. 10! is 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1, but since I'm dividing by 4!, which is 4 × 3 × 2 × 1, maybe I can simplify that.Wait, 10! divided by 4! is the same as 10 × 9 × 8 × 7 × 6 × 5. Because 10! is 10 × 9 × 8 × 7 × 6 × 5 × 4!, so when I divide by 4!, those terms cancel out. So, 10!/4! = 10 × 9 × 8 × 7 × 6 × 5.Let me compute that step by step:10 × 9 = 9090 × 8 = 720720 × 7 = 50405040 × 6 = 3024030240 × 5 = 151200So, 10!/4! is 151200.Now, I need to find the GCD of 5040 and 151200. Hmm, okay. One way to find the GCD is to use the prime factorization method. Let me try that.First, let me factorize 5040 into its prime factors.5040: I know that 5040 is 7! which I already calculated. Let me break it down:5040 ÷ 2 = 25202520 ÷ 2 = 12601260 ÷ 2 = 630630 ÷ 2 = 315315 ÷ 3 = 105105 ÷ 3 = 3535 ÷ 5 = 77 ÷ 7 = 1So, the prime factors are: 2^4 × 3^2 × 5 × 7.Now, let me factorize 151200.151200: Let's divide by 2 first.151200 ÷ 2 = 7560075600 ÷ 2 = 3780037800 ÷ 2 = 1890018900 ÷ 2 = 94509450 ÷ 2 = 4725Now, 4725 is not divisible by 2. Let's try 3.4725 ÷ 3 = 15751575 ÷ 3 = 525525 ÷ 3 = 175175 ÷ 5 = 3535 ÷ 5 = 77 ÷ 7 = 1So, the prime factors are: 2^5 × 3^3 × 5^2 × 7.Now, to find the GCD, I need to take the minimum exponents of the common prime factors.Looking at both factorizations:- For 2: 4 vs. 5 → take 4- For 3: 2 vs. 3 → take 2- For 5: 1 vs. 2 → take 1- For 7: 1 vs. 1 → take 1So, the GCD is 2^4 × 3^2 × 5 × 7.Let me compute that:2^4 = 163^2 = 916 × 9 = 144144 × 5 = 720720 × 7 = 5040Wait, that's interesting. The GCD is 5040, which is the same as 7!. That makes sense because 7! is a factor of 10!/4! since 10! includes 7! and we're dividing by 4!, which doesn't affect the 7! part. So, 7! divides 10!/4!, meaning the GCD is 7!.But just to make sure I didn't make a mistake, let me double-check.I found that 7! is 5040 and 10!/4! is 151200. The prime factors of 5040 are 2^4 × 3^2 × 5 × 7, and the prime factors of 151200 are 2^5 × 3^3 × 5^2 × 7. The GCD is the product of the minimum exponents of the common primes, which gives 2^4 × 3^2 × 5 × 7 = 5040. So, yes, that seems correct.Alternatively, I could use the Euclidean algorithm to find the GCD. Let me try that method too to confirm.The Euclidean algorithm says that GCD(a, b) = GCD(b, a mod b). So, let's apply that to 151200 and 5040.First, compute 151200 ÷ 5040. Let's see, 5040 × 30 = 151200. So, 151200 ÷ 5040 = 30 with a remainder of 0.Since the remainder is 0, the GCD is the smaller number, which is 5040. So, that confirms it again.Therefore, the greatest common divisor of 7! and 10!/4! is 5040.

💡First, I need to determine the distance the tiger has traveled during its 5-hour head start. Given that the tiger's average speed is 30 km/h, the distance covered in 5 hours is:5 hours × 30 km/h = 150 kmNext, I'll calculate the total distance the tiger travels by the time the zebra catches up. The zebra takes 6 hours to catch up, so during this time, the tiger travels an additional:6 hours × 30 km/h = 180 kmAdding this to the initial 150 km, the total distance the tiger has traveled when caught is:150 km + 180 km = 330 kmNow, I'll set up an equation to find the zebra's average speed. The zebra covers 330 km in 6 hours, so:6 hours × V_z = 330 kmSolving for V_z:V_z = 330 km / 6 hours = 55 km/hTherefore, the average speed of the zebra is 55 km/h.

💡Okay, so I have this problem: I need to find all the perfect squares that, when a given integer ( a ) is added to them, result in another perfect square. Hmm, that sounds interesting. Let me try to wrap my head around it.First, let's break down what the problem is asking. We have a perfect square, let's call it ( x^2 ). When we add some integer ( a ) to it, we get another perfect square, say ( y^2 ). So, mathematically, this can be written as:[ x^2 + a = y^2 ]Alright, so I need to find all such ( x^2 ) that satisfy this equation for a given ( a ). Let me see how I can approach this.Maybe I can rearrange the equation to make it easier to work with. If I subtract ( x^2 ) from both sides, I get:[ y^2 - x^2 = a ]Oh, wait! That looks like a difference of squares. I remember that ( y^2 - x^2 ) can be factored into ( (y - x)(y + x) ). So, substituting that in, the equation becomes:[ (y - x)(y + x) = a ]Alright, so now I have the product of two numbers, ( (y - x) ) and ( (y + x) ), equal to ( a ). This seems useful because it means I can think about the factors of ( a ) and set them equal to ( (y - x) ) and ( (y + x) ).Let me denote ( (y - x) = m ) and ( (y + x) = n ). So, ( m times n = a ). That means ( m ) and ( n ) are factors of ( a ). Also, since ( y ) and ( x ) are integers, ( m ) and ( n ) must also be integers.Now, since ( m = y - x ) and ( n = y + x ), I can solve for ( y ) and ( x ) in terms of ( m ) and ( n ). Let's add the two equations:[ m + n = (y - x) + (y + x) = 2y ][ Rightarrow y = frac{m + n}{2} ]Similarly, subtracting the two equations:[ n - m = (y + x) - (y - x) = 2x ][ Rightarrow x = frac{n - m}{2} ]So, both ( y ) and ( x ) must be integers, which means that ( m + n ) and ( n - m ) must both be even numbers. Therefore, ( m ) and ( n ) must be both even or both odd. In other words, ( m ) and ( n ) must have the same parity.This is an important point because if ( m ) and ( n ) have different parities, ( x ) and ( y ) won't be integers, which doesn't make sense in this context since we're dealing with perfect squares.Now, let's think about how to find all such pairs ( (m, n) ) such that ( m times n = a ) and ( m ) and ( n ) have the same parity. Essentially, I need to list all the factor pairs of ( a ) and check their parity.But before I proceed, I should consider the nature of ( a ). If ( a ) is positive, negative, or zero? Well, perfect squares are non-negative, so ( x^2 ) and ( y^2 ) are both non-negative. Therefore, ( a ) must be such that ( y^2 ) is greater than or equal to ( x^2 ). So, ( a ) must be non-negative as well. Wait, actually, ( a ) could be negative if ( y^2 < x^2 ), but since both are perfect squares, ( y^2 ) would have to be less than ( x^2 ), making ( a ) negative. But the problem says "a given integer ( a )", so ( a ) could be positive or negative.Hmm, but if ( a ) is negative, then ( y^2 = x^2 + a ) would mean ( y^2 < x^2 ). So, ( y ) would be less than ( x ). But since ( y ) and ( x ) are both non-negative integers (since they are squares), this is possible. So, ( a ) can be positive or negative.But let's stick to positive ( a ) for now to simplify things, and maybe later consider negative ( a ).So, assuming ( a ) is positive, I need to find all factor pairs ( (m, n) ) such that ( m times n = a ) and ( m ) and ( n ) have the same parity.Let me take an example to make this concrete. Suppose ( a = 15 ). Then, the factor pairs of 15 are (1,15) and (3,5). Let's check their parities.For (1,15): 1 is odd, 15 is odd. Same parity. Good.For (3,5): 3 is odd, 5 is odd. Same parity. Good.So, both factor pairs have the same parity. Therefore, both can be used to find ( x ) and ( y ).Using the first pair (1,15):[ y = frac{1 + 15}{2} = 8 ][ x = frac{15 - 1}{2} = 7 ]So, ( x^2 = 49 ) and ( y^2 = 64 ). Indeed, ( 49 + 15 = 64 ).Using the second pair (3,5):[ y = frac{3 + 5}{2} = 4 ][ x = frac{5 - 3}{2} = 1 ]So, ( x^2 = 1 ) and ( y^2 = 16 ). Indeed, ( 1 + 15 = 16 ).So, for ( a = 15 ), there are two perfect squares: 1 and 49.Wait, but ( x^2 ) is 1 and 49. So, the perfect squares that satisfy the condition are 1 and 49.But let's try another example where ( a ) is even. Let's take ( a = 16 ).Factor pairs of 16 are (1,16), (2,8), (4,4).Check their parities:(1,16): 1 is odd, 16 is even. Different parity. Not good.(2,8): Both even. Good.(4,4): Both even. Good.So, only the factor pairs (2,8) and (4,4) can be used.Using (2,8):[ y = frac{2 + 8}{2} = 5 ][ x = frac{8 - 2}{2} = 3 ]So, ( x^2 = 9 ), ( y^2 = 25 ). Indeed, ( 9 + 16 = 25 ).Using (4,4):[ y = frac{4 + 4}{2} = 4 ][ x = frac{4 - 4}{2} = 0 ]So, ( x^2 = 0 ), ( y^2 = 16 ). Indeed, ( 0 + 16 = 16 ).So, for ( a = 16 ), the perfect squares are 0 and 9.Wait, but 0 is a perfect square, right? ( 0 = 0^2 ). So, that's valid.But in the first example, ( a = 15 ), we had two perfect squares: 1 and 49. For ( a = 16 ), we have 0 and 9.Hmm, interesting. So, depending on the factor pairs, we get different perfect squares.Now, let's consider a negative ( a ). Let's say ( a = -7 ).So, ( x^2 + (-7) = y^2 ), which implies ( y^2 = x^2 - 7 ).So, ( y^2 ) must be less than ( x^2 ). Let's see if this is possible.We need to find integers ( x ) and ( y ) such that ( y^2 = x^2 - 7 ).Let's try small values of ( x ):If ( x = 3 ), ( x^2 = 9 ). Then ( y^2 = 9 - 7 = 2 ). But 2 is not a perfect square.If ( x = 4 ), ( x^2 = 16 ). Then ( y^2 = 16 - 7 = 9 ). 9 is a perfect square, ( y = 3 ).So, ( x = 4 ), ( y = 3 ). Therefore, ( x^2 = 16 ) is a perfect square that when ( a = -7 ) is added, results in ( y^2 = 9 ).Is there another solution?If ( x = 5 ), ( x^2 = 25 ). Then ( y^2 = 25 - 7 = 18 ). Not a perfect square.( x = 6 ), ( y^2 = 36 - 7 = 29 ). Not a perfect square.( x = 2 ), ( y^2 = 4 - 7 = -3 ). Negative, so not a perfect square.( x = 1 ), ( y^2 = 1 - 7 = -6 ). Negative.( x = 0 ), ( y^2 = 0 - 7 = -7 ). Negative.So, the only solution is ( x^2 = 16 ).Wait, but according to our earlier method, let's see if we can apply it here.We have ( y^2 - x^2 = -7 ), so ( (y - x)(y + x) = -7 ).So, factor pairs of -7 are (-1,7), (1,-7), (-7,1), (7,-1).Let's take each pair:1. ( m = -1 ), ( n = 7 ):[ y = frac{-1 + 7}{2} = 3 ][ x = frac{7 - (-1)}{2} = 4 ]So, ( x = 4 ), ( y = 3 ). That works.2. ( m = 1 ), ( n = -7 ):[ y = frac{1 + (-7)}{2} = -3 ][ x = frac{-7 - 1}{2} = -4 ]But since we're dealing with perfect squares, ( x ) and ( y ) can be negative, but their squares are still positive. So, ( x = -4 ), ( y = -3 ) also works, but it's essentially the same solution as ( x = 4 ), ( y = 3 ).3. ( m = -7 ), ( n = 1 ):[ y = frac{-7 + 1}{2} = -3 ][ x = frac{1 - (-7)}{2} = 4 ]Again, same solution.4. ( m = 7 ), ( n = -1 ):[ y = frac{7 + (-1)}{2} = 3 ][ x = frac{-1 - 7}{2} = -4 ]Same as before.So, essentially, only one unique solution: ( x = 4 ), ( y = 3 ), giving ( x^2 = 16 ).So, that works.But in this case, the factor pairs had different signs, but still, we could get integer solutions because the sum and difference ended up being even.Wait, but in this case, ( m = -1 ), ( n = 7 ). Their sum is 6, which is even, and their difference is 8, which is also even. So, ( y ) and ( x ) are integers.But if I take another factor pair, say ( m = -3 ), ( n = something ), but wait, -7 only has factors -1, 1, -7, 7.So, in this case, only one unique solution.So, seems like the method works for negative ( a ) as well.But let's think about ( a = 0 ). If ( a = 0 ), then ( x^2 + 0 = y^2 ), so ( x^2 = y^2 ). Therefore, ( x = y ) or ( x = -y ). But since we're dealing with perfect squares, which are non-negative, ( x = y ).So, for ( a = 0 ), every perfect square ( x^2 ) satisfies the condition because adding zero doesn't change it.But the problem says "a given integer ( a )", so ( a ) could be zero, but in that case, all perfect squares satisfy the condition trivially.But I think the problem is more interesting when ( a ) is non-zero.So, going back, the general approach is:1. Write ( x^2 + a = y^2 ).2. Rearrange to ( y^2 - x^2 = a ).3. Factor as ( (y - x)(y + x) = a ).4. Find all factor pairs ( (m, n) ) of ( a ) such that ( m times n = a ) and ( m ) and ( n ) have the same parity.5. For each such pair, solve for ( x ) and ( y ) using ( x = frac{n - m}{2} ) and ( y = frac{n + m}{2} ).6. Ensure that ( x ) and ( y ) are integers, which they will be if ( m ) and ( n ) have the same parity.So, this seems to be a systematic way to find all such perfect squares ( x^2 ) for a given ( a ).But let's think about the cases where ( a ) is even or odd.If ( a ) is odd, then all its factor pairs will consist of two odd numbers because an odd number times an odd number is odd. So, in that case, all factor pairs will have the same parity (both odd), so all factor pairs will work.For example, ( a = 15 ) is odd, and both factor pairs (1,15) and (3,5) are odd, so both give valid solutions.But if ( a ) is even, then its factor pairs can be both even or one even and one odd. But since ( m ) and ( n ) must have the same parity, only the factor pairs where both are even will work.For example, ( a = 16 ) is even. Its factor pairs are (1,16), (2,8), (4,4). Among these, (1,16) has different parity, so it doesn't work, but (2,8) and (4,4) are both even, so they work.Wait, but what if ( a ) is even but not divisible by 4? Let's take ( a = 6 ).Factor pairs of 6 are (1,6), (2,3).Check their parities:(1,6): 1 is odd, 6 is even. Different parity. Doesn't work.(2,3): 2 is even, 3 is odd. Different parity. Doesn't work.So, no solutions for ( a = 6 ).Wait, but is that true? Let's check manually.We need ( x^2 + 6 = y^2 ).Trying small ( x ):( x = 1 ): ( 1 + 6 = 7 ). Not a square.( x = 2 ): ( 4 + 6 = 10 ). Not a square.( x = 3 ): ( 9 + 6 = 15 ). Not a square.( x = 4 ): ( 16 + 6 = 22 ). Not a square.( x = 5 ): ( 25 + 6 = 31 ). Not a square.( x = 0 ): ( 0 + 6 = 6 ). Not a square.So, indeed, no solutions for ( a = 6 ).So, if ( a ) is even but not divisible by 4, there are no solutions. Because in that case, all factor pairs will consist of one even and one odd number, which have different parities, so no valid ( x ) and ( y ).But if ( a ) is divisible by 4, then it's possible to have factor pairs where both are even.For example, ( a = 4 ).Factor pairs: (1,4), (2,2).(1,4): Different parity. Doesn't work.(2,2): Both even. Works.So, using (2,2):[ y = frac{2 + 2}{2} = 2 ][ x = frac{2 - 2}{2} = 0 ]So, ( x^2 = 0 ), ( y^2 = 4 ). Indeed, ( 0 + 4 = 4 ).Another example, ( a = 8 ).Factor pairs: (1,8), (2,4).(1,8): Different parity. Doesn't work.(2,4): Both even. Works.So, using (2,4):[ y = frac{2 + 4}{2} = 3 ][ x = frac{4 - 2}{2} = 1 ]So, ( x^2 = 1 ), ( y^2 = 9 ). Indeed, ( 1 + 8 = 9 ).So, in this case, there is one solution.Wait, but ( a = 8 ) is divisible by 4? No, 8 divided by 4 is 2, so yes, it's divisible by 4.Wait, 8 is 2^3, so it's divisible by 4.But in the case of ( a = 6 ), which is 2 × 3, not divisible by 4, there are no solutions.So, seems like if ( a ) is even, it must be divisible by 4 to have solutions.Wait, let me check ( a = 12 ).Factor pairs: (1,12), (2,6), (3,4).Check parities:(1,12): Different parity.(2,6): Both even.(3,4): Different parity.So, only (2,6) works.Using (2,6):[ y = frac{2 + 6}{2} = 4 ][ x = frac{6 - 2}{2} = 2 ]So, ( x^2 = 4 ), ( y^2 = 16 ). Indeed, ( 4 + 12 = 16 ).Another factor pair for ( a = 12 ) is (3,4), but they have different parity, so it doesn't work.So, only one solution.But ( a = 12 ) is divisible by 4? 12 divided by 4 is 3, so yes, it's divisible by 4.Wait, but 12 is 4 × 3, so it's divisible by 4.So, seems like if ( a ) is even, it must be divisible by 4 to have solutions.But wait, ( a = 4 ) is divisible by 4, and we had a solution.( a = 8 ) is divisible by 4, and we had a solution.( a = 12 ) is divisible by 4, and we had a solution.But ( a = 16 ) is divisible by 4, and we had two solutions.Wait, but ( a = 16 ) is 4^2, so it's a perfect square itself.Hmm, interesting.So, in summary:- If ( a ) is odd, then it can be expressed as a difference of two squares in multiple ways, depending on its factor pairs.- If ( a ) is even, it must be divisible by 4 to be expressible as a difference of two squares. If it's divisible by 4, then it can be expressed as a difference of two squares.But wait, is that always true?Let me think about ( a = 20 ).Factor pairs: (1,20), (2,10), (4,5).Check parities:(1,20): Different parity.(2,10): Both even.(4,5): Different parity.So, only (2,10) works.Using (2,10):[ y = frac{2 + 10}{2} = 6 ][ x = frac{10 - 2}{2} = 4 ]So, ( x^2 = 16 ), ( y^2 = 36 ). Indeed, ( 16 + 20 = 36 ).Another factor pair for ( a = 20 ) is (4,5), but different parity, so no solution.So, only one solution.But ( a = 20 ) is divisible by 4, as 20 = 4 × 5.So, seems consistent.But what about ( a = 24 ).Factor pairs: (1,24), (2,12), (3,8), (4,6).Check parities:(1,24): Different parity.(2,12): Both even.(3,8): Different parity.(4,6): Both even.So, two factor pairs: (2,12) and (4,6).Using (2,12):[ y = frac{2 + 12}{2} = 7 ][ x = frac{12 - 2}{2} = 5 ]So, ( x^2 = 25 ), ( y^2 = 49 ). Indeed, ( 25 + 24 = 49 ).Using (4,6):[ y = frac{4 + 6}{2} = 5 ][ x = frac{6 - 4}{2} = 1 ]So, ( x^2 = 1 ), ( y^2 = 25 ). Indeed, ( 1 + 24 = 25 ).So, two solutions for ( a = 24 ).So, seems like when ( a ) is divisible by 4, the number of solutions depends on the number of factor pairs where both factors are even.So, in general, for a given ( a ), the number of perfect squares ( x^2 ) such that ( x^2 + a = y^2 ) is equal to the number of factor pairs ( (m, n) ) of ( a ) where ( m ) and ( n ) have the same parity.But to find all such ( x^2 ), I need to:1. List all factor pairs ( (m, n) ) of ( a ).2. For each pair, check if ( m ) and ( n ) have the same parity.3. For each valid pair, compute ( x = frac{n - m}{2} ) and ( y = frac{n + m}{2} ).4. Collect all such ( x^2 ).So, that's the method.But let's think about how to implement this or how to describe it.First, factorizing ( a ) into all possible pairs ( (m, n) ) such that ( m times n = a ).But factorizing can be time-consuming for large ( a ), but since we're just looking for a method, it's acceptable.Now, considering that ( m ) and ( n ) must have the same parity, which depends on whether ( a ) is odd or even.If ( a ) is odd, all its factors are odd, so all factor pairs will have the same parity.If ( a ) is even, then ( a ) must be divisible by 4 for there to exist factor pairs with both even factors.Wait, is that necessarily true?Let me think.If ( a ) is even, it can be written as ( a = 2^k times b ), where ( b ) is odd.If ( k = 1 ), then ( a = 2 times b ), where ( b ) is odd.In this case, any factor pair ( (m, n) ) will have one even and one odd factor because ( a ) has only one factor of 2.Therefore, in this case, there are no factor pairs with both even factors, so no solutions.If ( k geq 2 ), then ( a = 4 times c ), where ( c ) is an integer.In this case, ( a ) has at least two factors of 2, so it's possible to have factor pairs where both ( m ) and ( n ) are even.Therefore, for ( a ) even, solutions exist only if ( a ) is divisible by 4.So, to summarize:- If ( a ) is odd, there are solutions, and the number of solutions depends on the number of factor pairs.- If ( a ) is even, solutions exist only if ( a ) is divisible by 4, and the number of solutions depends on the number of factor pairs where both factors are even.Therefore, the perfect squares ( x^2 ) that satisfy ( x^2 + a = y^2 ) are given by:For each factor pair ( (m, n) ) of ( a ) with ( m leq n ) and ( m ) and ( n ) having the same parity, compute ( x = frac{n - m}{2} ) and ( y = frac{n + m}{2} ). Then, ( x^2 ) is a perfect square that satisfies the condition.So, the final answer is that the perfect squares are those ( x^2 ) where ( x = frac{n - m}{2} ) for some factor pair ( (m, n) ) of ( a ) with ( m ) and ( n ) having the same parity.But to write this in a box as requested, I think the answer is that the perfect squares are given by ( x = frac{beta - alpha}{2} ) and ( y = frac{beta + alpha}{2} ) where ( alpha ) and ( beta ) are factors of ( a ) with the same parity.So, the boxed answer would be:boxed{x = frac{beta - alpha}{2}, y = frac{beta + alpha}{2}}But wait, the question is asking for the perfect squares, not ( x ) and ( y ). So, perhaps the answer should be the set of ( x^2 ) where ( x = frac{beta - alpha}{2} ) for some factor pair ( (alpha, beta) ) of ( a ) with the same parity.But since the question is asking for the perfect squares, not the values of ( x ), maybe I should express it differently.Alternatively, the perfect squares are those ( x^2 ) such that ( x = frac{beta - alpha}{2} ) where ( alpha ) and ( beta ) are factors of ( a ) with the same parity.But perhaps the answer is better expressed as the method to find them, rather than listing them.Alternatively, the perfect squares are given by ( x^2 = left( frac{beta - alpha}{2} right)^2 ) where ( alpha ) and ( beta ) are factors of ( a ) with the same parity.But I think the most concise way is to present the formula for ( x ) and ( y ), as that directly answers how to find the perfect squares.So, I think the boxed answer should be:boxed{x = frac{beta - alpha}{2}, y = frac{beta + alpha}{2}}where ( alpha ) and ( beta ) are factors of ( a ) with the same parity.But since the question is about the perfect squares, perhaps it's better to write the perfect squares as ( x^2 ) where ( x = frac{beta - alpha}{2} ).Alternatively, the perfect squares are ( left( frac{beta - alpha}{2} right)^2 ).But I think the original answer I thought of is acceptable, as it shows how to compute ( x ) and ( y ), which leads to the perfect squares.So, I'll stick with that.

💡First, I need to determine Louisa's average speed based on the distances she traveled and the time difference between the two trips.I'll start by defining her average speed as ( v ) miles per hour.On the first day, she traveled 200 miles, so the time taken for this trip is ( frac{200}{v} ) hours.On the second day, she traveled 350 miles, so the time taken for this trip is ( frac{350}{v} ) hours.According to the problem, the 200-mile trip took 3 hours less than the 350-mile trip. This gives me the equation:[frac{350}{v} - frac{200}{v} = 3]Simplifying the equation:[frac{150}{v} = 3]Solving for ( v ):[150 = 3v][v = frac{150}{3} = 50]Therefore, Louisa's average speed was 50 miles per hour.

💡First, I need to determine how many friends Joshua gave Skittles to. The total number of Skittles his friends have is 200, and each friend received 40 Skittles.To find the number of friends, I can divide the total number of Skittles by the number of Skittles each friend received.So, the number of friends is 200 divided by 40, which equals 5.Therefore, Joshua gave Skittles to 5 friends.