💡First, calculate the total cost of purchasing the 118 glass bowls.Next, determine the total selling price by adding the profit to the total cost.Finally, find the selling price per bowl by dividing the total selling price by the number of bowls sold.

💡Okay, so I came across this math problem that asks whether the least common multiple (LCM) of the integers from 1 to n can be 2008 times greater than the LCM of the integers from 1 to m. Hmm, that seems a bit tricky, but let me try to break it down.First, I remember that the LCM of a set of numbers is the smallest number that all of them divide into. So, LCM(1, 2, ..., n) is the smallest number that all numbers from 1 to n can divide into without leaving a remainder. Similarly, LCM(1, 2, ..., m) is the smallest number for the range up to m.The question is asking if the ratio of these two LCMs can be exactly 2008. That is, can LCM(1, 2, ..., n) / LCM(1, 2, ..., m) = 2008?I think the first step is to understand the properties of LCMs. I recall that LCM(1, 2, ..., n) is equal to the product of the highest powers of all primes less than or equal to n. For example, LCM(1, 2, 3, 4) is 12 because the primes involved are 2 and 3, and the highest powers are 2^2 and 3^1.So, if we have LCM(1, 2, ..., n) and LCM(1, 2, ..., m), their ratio would essentially be the product of the primes (and their powers) that are in the range m+1 to n. That makes sense because LCM(1, 2, ..., n) includes all the primes up to n, while LCM(1, 2, ..., m) only includes up to m. So, the ratio would be the product of the primes (and their higher powers) between m+1 and n.Given that, we need this product to be 2008. So, let's factorize 2008 to see what primes we're dealing with. 2008 divided by 2 is 1004, divided by 2 again is 502, and again by 2 is 251. 251 is a prime number. So, 2008 factors into 2^3 * 251.So, the ratio LCM(1, 2, ..., n) / LCM(1, 2, ..., m) must equal 2^3 * 251. That means that between m and n, the only primes contributing to the LCM ratio should be 2 and 251, with 2 raised to the third power and 251 raised to the first power.Wait, but 251 is a pretty large prime. So, if 251 is a prime between m+1 and n, then n has to be at least 251. Also, m has to be less than 251 because otherwise, 251 would already be included in LCM(1, 2, ..., m), and thus wouldn't contribute to the ratio.So, n must be at least 251, and m must be less than 251. Now, let's think about the power of 2. The ratio has 2^3, which means that in the LCM(1, 2, ..., n), the exponent of 2 must be exactly 3 more than in LCM(1, 2, ..., m).But wait, the exponent of 2 in LCM(1, 2, ..., k) is the largest integer such that 2^e ≤ k. So, if in LCM(1, 2, ..., n), the exponent of 2 is e_n, and in LCM(1, 2, ..., m), it's e_m, then e_n - e_m should be 3.So, we need that the exponent of 2 in LCM(1, 2, ..., n) is exactly 3 more than in LCM(1, 2, ..., m). That would mean that n is at least 2^(e_m + 3), but m is less than 2^(e_m + 1). Hmm, that might be a bit abstract. Let me try to make it more concrete.Suppose that m is just below a power of 2. Let's say m is between 2^k and 2^(k+1). Then, the exponent of 2 in LCM(1, 2, ..., m) is k. So, to get a difference of 3 in the exponents, n would need to be at least 2^(k+3). But wait, if n is 2^(k+3), then the exponent of 2 in LCM(1, 2, ..., n) would be k+3, which is exactly 3 more than k. So, that seems to fit.But we also have to consider the prime 251. So, n must be at least 251, and m must be less than 251. So, let's see. If m is less than 251, then the exponent of 2 in LCM(1, 2, ..., m) is the largest integer k such that 2^k ≤ m. Similarly, n must be at least 251, so the exponent of 2 in LCM(1, 2, ..., n) is the largest integer such that 2^k ≤ n.Now, if n is at least 251, what is the exponent of 2 in LCM(1, 2, ..., n)? Let's calculate 2^8 is 256, which is greater than 251, so the exponent of 2 would be 7 because 2^7 is 128, which is less than 251, and 2^8 is 256, which is greater. Wait, no, actually, the exponent is the largest integer k such that 2^k ≤ n. So, if n is 251, then 2^7 is 128, and 2^8 is 256, which is greater than 251. So, the exponent of 2 in LCM(1, 2, ..., 251) is 7.Similarly, if m is less than 251, say m is 127, which is 2^7 -1, then the exponent of 2 in LCM(1, 2, ..., 127) is 6 because 2^6 is 64, and 2^7 is 128, which is greater than 127. Wait, no, 2^7 is 128, which is greater than 127, so the exponent would be 6. Hmm, so if m is 127, the exponent of 2 is 6, and if n is 251, the exponent of 2 is 7. So, the difference is 1, not 3.But we need the difference to be 3. So, maybe m needs to be such that the exponent of 2 in LCM(1, 2, ..., m) is 4, and in LCM(1, 2, ..., n), it's 7. That would give a difference of 3. So, m would need to be less than 2^5=32, because the exponent of 2 would be 4 if m is less than 32. Then, n would need to be at least 2^7=128. But wait, n also needs to be at least 251 because of the prime 251.Wait, that doesn't make sense because if n is 251, the exponent of 2 is 7, and if m is less than 32, the exponent of 2 is 4, so the difference is 3. That seems to fit. But then, does the ratio LCM(1, 2, ..., 251)/LCM(1, 2, ..., m) equal 2^3 * 251?Wait, no, because LCM(1, 2, ..., 251) includes all primes up to 251, not just 2 and 251. So, if m is less than 251, say m is 31, then LCM(1, 2, ..., 31) doesn't include primes like 37, 41, etc., up to 251. So, the ratio would include all those primes as well, which would make the ratio much larger than 2008.Ah, that's a problem. So, the ratio isn't just 2^3 * 251, it's also multiplied by all the other primes between m+1 and n. So, unless those other primes somehow cancel out or are accounted for, the ratio would be larger than 2008.Wait, but 2008 is 2^3 * 251, so if there are other primes between m+1 and n, then the ratio would have additional prime factors, making it larger than 2008. Therefore, to have the ratio exactly equal to 2008, there should be no other primes between m+1 and n except for 251. But that's impossible because there are many primes between, say, 32 and 251.For example, primes like 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251. That's a lot of primes. So, unless m is just below 251 and n is 251, but then m would be 250, and LCM(1, 2, ..., 250) would already include all primes up to 250, so the ratio would be 251, not 2008.Wait, but 2008 is 8 * 251, so maybe if n is 251 and m is such that the exponent of 2 in LCM(1, 2, ..., m) is 4 less than in LCM(1, 2, ..., 251). Wait, no, the exponent difference is 3, as we saw earlier. So, if the exponent of 2 in LCM(1, 2, ..., m) is 4, then m is less than 32, and n is 251, which has exponent 7. So, the ratio would be 2^(7-4) * 251, which is 8 * 251 = 2008. But wait, does that mean that the ratio is exactly 2008?But hold on, LCM(1, 2, ..., 251) includes all primes up to 251, and LCM(1, 2, ..., m) includes all primes up to m. So, if m is less than 251, then the ratio would include all primes between m+1 and 251. But 2008 is only 2^3 * 251, so unless all those other primes between m+1 and 251 are somehow accounted for, the ratio would be larger.Wait, that can't be. Because if m is less than 251, say m is 31, then LCM(1, 2, ..., 31) doesn't include primes like 37, 41, etc., up to 251. So, the ratio would be LCM(1, 2, ..., 251) / LCM(1, 2, ..., 31) = (2^7 * 3^3 * 5^2 * 7^2 * 11 * 13 * 17 * 19 * 23 * 29 * 31 * ... * 251) / (2^4 * 3^3 * 5^2 * 7^2 * 11 * 13 * 17 * 19 * 23 * 29 * 31). Wait, no, actually, LCM(1, 2, ..., 31) includes primes up to 31, but LCM(1, 2, ..., 251) includes primes up to 251. So, the ratio would be the product of all primes between 32 and 251, each raised to their respective exponents, multiplied by the increase in exponents of existing primes.But 2008 is just 2^3 * 251. So, unless all those other primes between 32 and 251 somehow don't contribute, which they do, the ratio would be much larger than 2008. Therefore, it's impossible for the ratio to be exactly 2008 because it would require that there are no other primes between m+1 and n except for 251, which isn't the case.Wait, but maybe if m is just below 251, like m=250, then LCM(1, 2, ..., 250) already includes all primes up to 250, so the ratio would be 251, not 2008. So, that doesn't work either.Alternatively, maybe if m is such that the exponent of 2 in LCM(1, 2, ..., m) is 4, and n is such that the exponent of 2 is 7, and the only new prime is 251. But as we saw, there are many primes between m+1 and n, so the ratio would include all of them, making it larger than 2008.Therefore, it seems that it's impossible for the ratio to be exactly 2008 because the ratio would necessarily include other prime factors beyond 2 and 251, making it larger than 2008.Wait, but let me double-check. Suppose m is 127, which is 2^7 -1. Then, LCM(1, 2, ..., 127) has exponent of 2 as 6 because 2^7 is 128, which is greater than 127. Then, n needs to be at least 2^(6+3)=512. But 512 is much larger than 251, so LCM(1, 2, ..., 512) includes all primes up to 512, which includes 251 and many others. So, the ratio would be 2^3 * (product of all primes between 128 and 512). That's way more than 2008.Alternatively, if m is 250, then LCM(1, 2, ..., 250) includes all primes up to 250, and n is 251. Then, the ratio is 251, which is less than 2008. So, that doesn't work.Wait, maybe if m is such that the exponent of 2 in LCM(1, 2, ..., m) is 4, and n is such that the exponent of 2 is 7, and the only new prime is 251. But as we saw, there are many primes between m+1 and n, so the ratio would include all of them, making it larger than 2008.Therefore, it seems that it's impossible for the ratio to be exactly 2008 because the ratio would necessarily include other prime factors beyond 2 and 251, making it larger than 2008.Wait, but let me think again. Maybe if m is chosen such that the only new prime between m and n is 251, and the exponent of 2 increases by 3. But that would require that there are no other primes between m and n except 251, which is impossible because there are many primes between, say, 128 and 251.Therefore, the answer should be that it's not possible.

💡First, I need to determine the smallest positive integer ( n ) such that ( n^2 ) is greater than 50. Calculating the square root of 50 gives approximately 7.07, so the smallest integer ( n ) is 8 because ( 8^2 = 64 ).Next, I need to find the largest positive integer ( n ) such that ( n^2 ) is less than 900. The square root of 900 is exactly 30, but since ( n^2 ) must be less than 900, the largest integer ( n ) is 29 because ( 29^2 = 841 ).Finally, I count the integers from 8 to 29. This range includes 22 integers. Therefore, there are 22 positive integers ( n ) that satisfy the inequality ( 50 < n^2 < 900 ).

💡First, I need to identify the unknown number, which I'll call ( x ).The given equation is:[frac{3.6 times 0.48 times x}{0.12 times 0.09 times 0.5} = 800.0000000000001]Next, I'll calculate the product of the constants in the numerator:[3.6 times 0.48 = 1.728]Then, I'll calculate the product of the constants in the denominator:[0.12 times 0.09 times 0.5 = 0.0054]Now, the equation simplifies to:[frac{1.728 times x}{0.0054} = 800.0000000000001]To isolate ( x ), I'll multiply both sides of the equation by ( 0.0054 ):[1.728 times x = 800.0000000000001 times 0.0054]Calculating the right side:[800.0000000000001 times 0.0054 = 4.32]Now, the equation is:[1.728 times x = 4.32]Finally, I'll solve for ( x ) by dividing both sides by ( 1.728 ):[x = frac{4.32}{1.728} = 2.5]Therefore, the third number is ( 2.5 ).

💡Okay, so I have this problem about a function ( f(x) = frac{a^x + a - 3}{ln a} ) where ( a > 0 ) and ( a neq 1 ). The problem has two parts: first, determining the monotonicity of the function on ( (-infty, +infty) ), and second, finding the range of ( a ) such that the function is injective over the real numbers ( mathbb{R} ).Starting with part (1), I need to figure out if the function is increasing, decreasing, or something else. Since the function involves an exponential term ( a^x ), I remember that the derivative of ( a^x ) with respect to ( x ) is ( a^x ln a ). But in this case, the function is divided by ( ln a ), so maybe that simplifies things.Let me compute the derivative of ( f(x) ). The function is ( f(x) = frac{a^x + a - 3}{ln a} ). So, the derivative ( f'(x) ) would be the derivative of the numerator divided by ( ln a ). The derivative of ( a^x ) is ( a^x ln a ), and the derivatives of ( a ) and ( -3 ) are zero. Therefore, ( f'(x) = frac{a^x ln a}{ln a} = a^x ).Since ( a > 0 ) and ( a neq 1 ), ( a^x ) is always positive for all real ( x ). Therefore, ( f'(x) = a^x > 0 ) for all ( x ). This means the function is strictly increasing on ( (-infty, +infty) ). So, the answer to part (1) is that the function is increasing.Moving on to part (2), the function is called injective in ( D ) if it's monotonic and there exists an interval ( [m, n] ) such that the range of ( f ) over ( [m, n] ) is also ( [m, n] ). Since we already know the function is monotonic (specifically, increasing), we just need to ensure that there exists such an interval ( [m, n] ).For the function to be injective, it must satisfy ( f(m) = m ) and ( f(n) = n ). That is, ( m ) and ( n ) are fixed points of the function. So, we need to solve the equation ( f(x) = x ) for ( x ). Let's set up that equation:( frac{a^x + a - 3}{ln a} = x )Multiplying both sides by ( ln a ) gives:( a^x + a - 3 = x ln a )Rearranging terms:( a^x - x ln a + a - 3 = 0 )Let me define a new function ( g(x) = a^x - x ln a + a - 3 ). We need to find the number of real roots of ( g(x) = 0 ). Since the function is injective, we need exactly two real roots ( m ) and ( n ) such that ( m < n ) and ( f([m, n]) = [m, n] ).To analyze the roots, let's look at the derivative of ( g(x) ):( g'(x) = a^x ln a - ln a = ln a (a^x - 1) )The critical points occur where ( g'(x) = 0 ), which is when ( a^x - 1 = 0 ) since ( ln a neq 0 ) (because ( a neq 1 )). So, ( a^x = 1 ) implies ( x = 0 ).Therefore, ( x = 0 ) is the critical point. Let's analyze the behavior of ( g(x) ) around this point.Case 1: ( a > 1 )In this case, ( ln a > 0 ). So, for ( x < 0 ), ( a^x < 1 ), which means ( g'(x) = ln a (a^x - 1) < 0 ). For ( x > 0 ), ( a^x > 1 ), so ( g'(x) > 0 ). Therefore, ( g(x) ) is decreasing on ( (-infty, 0) ) and increasing on ( (0, +infty) ). Thus, ( x = 0 ) is a minimum point.To have two real roots, the function must cross the x-axis twice. So, the minimum value at ( x = 0 ) must be below zero. Let's compute ( g(0) ):( g(0) = a^0 - 0 cdot ln a + a - 3 = 1 + a - 3 = a - 2 )For ( g(0) < 0 ), we need ( a - 2 < 0 ) which implies ( a < 2 ). Since we're in the case ( a > 1 ), combining these gives ( 1 < a < 2 ).Case 2: ( 0 < a < 1 )Here, ( ln a < 0 ). For ( x < 0 ), ( a^x > 1 ) because ( a < 1 ) and raising it to a negative power makes it greater than 1. So, ( g'(x) = ln a (a^x - 1) ). Since ( ln a < 0 ) and ( a^x - 1 > 0 ), ( g'(x) < 0 ). For ( x > 0 ), ( a^x < 1 ), so ( a^x - 1 < 0 ), and since ( ln a < 0 ), ( g'(x) = ln a (a^x - 1) > 0 ). Therefore, ( g(x) ) is decreasing on ( (-infty, 0) ) and increasing on ( (0, +infty) ), similar to the previous case. Thus, ( x = 0 ) is still a minimum point.Again, to have two real roots, the minimum at ( x = 0 ) must be below zero. Compute ( g(0) ):( g(0) = a^0 - 0 cdot ln a + a - 3 = 1 + a - 3 = a - 2 )For ( g(0) < 0 ), ( a < 2 ). But since ( 0 < a < 1 ), this condition is automatically satisfied because ( a ) is already less than 1, which is less than 2. Therefore, for ( 0 < a < 1 ), ( g(0) = a - 2 < 0 ), so the function ( g(x) ) will have two real roots.Putting it all together, the function ( f(x) ) is injective when ( 0 < a < 1 ) or ( 1 < a < 2 ). Therefore, the range of ( a ) is ( (0, 1) cup (1, 2) ).I think that covers both parts of the problem. For part (1), the function is increasing everywhere, and for part (2), the range of ( a ) is between 0 and 1, excluding 1, and between 1 and 2, excluding 1 again.**Final Answer**(1) boxed{text{Increasing}}(2) boxed{(0, 1) cup (1, 2)}

💡Alright, so I have this problem here that I need to prove: for every nonnegative integer ( n ), there exists an odd number ( m ) and a parabolic triangle with vertices at three distinct points with integer coordinates such that the area of the triangle is ( (2^n m)^2 ). Hmm, okay, let's break this down.First off, what exactly is a parabolic triangle? I think it means a triangle whose vertices lie on a parabola. So, if we consider the standard parabola ( y = x^2 ), the vertices of the triangle would be points on this curve. That makes sense. So, the vertices would be of the form ( (a, a^2) ), ( (b, b^2) ), and ( (c, c^2) ) where ( a ), ( b ), and ( c ) are integers.Now, the area of a triangle given three points can be calculated using the shoelace formula or the determinant method. Let me recall the formula. For three points ( (x_1, y_1) ), ( (x_2, y_2) ), and ( (x_3, y_3) ), the area is:[text{Area} = frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|]So, plugging in the points ( (a, a^2) ), ( (b, b^2) ), and ( (c, c^2) ), the area becomes:[text{Area} = frac{1}{2} |a(b^2 - c^2) + b(c^2 - a^2) + c(a^2 - b^2)|]Simplify that a bit:[text{Area} = frac{1}{2} |a(b^2 - c^2) + b(c^2 - a^2) + c(a^2 - b^2)|]Let me factor some terms here. Notice that ( b^2 - c^2 = (b - c)(b + c) ), and similarly for the other terms. So, expanding each term:[= frac{1}{2} |a(b - c)(b + c) + b(c - a)(c + a) + c(a - b)(a + b)|]Hmm, this seems a bit complicated. Maybe there's a better way to approach this. Alternatively, I remember that for three points on a parabola, the area can also be expressed in terms of the differences of the x-coordinates. Maybe I can find a formula that relates the area directly to ( a ), ( b ), and ( c ).Wait, another thought: if I choose two points symmetric about the y-axis, say ( (a, a^2) ) and ( (-a, a^2) ), then the base of the triangle would be ( 2a ). The third point can be somewhere else on the parabola, say ( (b, b^2) ). Then, the height of the triangle would be the vertical distance from ( (b, b^2) ) to the line connecting ( (a, a^2) ) and ( (-a, a^2) ).But wait, the line connecting ( (a, a^2) ) and ( (-a, a^2) ) is horizontal, right? Because both points have the same y-coordinate ( a^2 ). So, the height of the triangle would just be ( |b^2 - a^2| ).Therefore, the area of the triangle would be:[text{Area} = frac{1}{2} times text{base} times text{height} = frac{1}{2} times 2a times |b^2 - a^2| = a |b^2 - a^2|]Okay, that's a much simpler expression! So, if I choose the vertices as ( (a, a^2) ), ( (-a, a^2) ), and ( (b, b^2) ), the area is ( a |b^2 - a^2| ).Now, the problem wants the area to be ( (2^n m)^2 ), where ( m ) is an odd integer. So, we have:[a |b^2 - a^2| = (2^n m)^2]Our goal is to find integers ( a ), ( b ), and ( m ) such that this equation holds, with ( m ) being odd.Let me think about how to choose ( a ) and ( b ) such that this equation is satisfied. Since ( a ) and ( b ) are integers, ( |b^2 - a^2| ) must be a positive integer as well. Also, ( a ) must divide ( (2^n m)^2 ).Since ( m ) is odd, ( (2^n m)^2 = 2^{2n} m^2 ). So, ( a ) must be a divisor of ( 2^{2n} m^2 ). Let's choose ( a ) as a power of 2 to simplify things. Let me set ( a = 2^n ). Then, the equation becomes:[2^n |b^2 - (2^n)^2| = 2^{2n} m^2]Divide both sides by ( 2^n ):[|b^2 - 2^{2n}| = 2^n m^2]So, we have:[b^2 - 2^{2n} = pm 2^n m^2]But since ( b^2 ) is positive and greater than ( 2^{2n} ) (assuming ( b > 2^n )), we can drop the absolute value:[b^2 - 2^{2n} = 2^n m^2]So, we need to find integers ( b ) and ( m ) such that:[b^2 - 2^{2n} = 2^n m^2]Let me rearrange this:[b^2 = 2^{2n} + 2^n m^2]Factor out ( 2^n ):[b^2 = 2^n (2^n + m^2)]Since ( b^2 ) is a perfect square, the right-hand side must also be a perfect square. Let me denote ( 2^n + m^2 = k^2 ), where ( k ) is an integer. Then, we have:[b^2 = 2^n k^2]Which implies:[b = k sqrt{2^n}]But ( b ) must be an integer, so ( sqrt{2^n} ) must be rational, which it is only if ( n ) is even. Wait, but ( n ) can be any nonnegative integer, including odd numbers. Hmm, this seems problematic.Wait, maybe my approach is wrong. Let me think again.We have:[b^2 = 2^n (2^n + m^2)]For ( b ) to be an integer, ( 2^n (2^n + m^2) ) must be a perfect square. Let me denote ( 2^n + m^2 = 2^{n} t^2 ), where ( t ) is an integer. Then:[b^2 = 2^n times 2^n t^2 = 2^{2n} t^2][b = 2^n t]So, substituting back:[2^n + m^2 = 2^n t^2][m^2 = 2^n (t^2 - 1)]Since ( m ) is odd, ( m^2 ) is also odd. Therefore, the right-hand side ( 2^n (t^2 - 1) ) must be odd. But ( 2^n ) is a power of 2, which is even unless ( n = 0 ). So, for ( n geq 1 ), ( 2^n ) is even, and ( t^2 - 1 ) must be odd to make the product odd. But ( t^2 - 1 ) is always even because ( t^2 ) is either even or odd:- If ( t ) is even, ( t^2 ) is even, so ( t^2 - 1 ) is odd.- If ( t ) is odd, ( t^2 ) is odd, so ( t^2 - 1 ) is even.Wait, so if ( t ) is even, ( t^2 - 1 ) is odd, which would make ( m^2 = 2^n times text{odd} ). But ( m^2 ) is odd, so ( 2^n ) must be 1, meaning ( n = 0 ). But we are considering ( n geq 1 ). This seems like a contradiction.Hmm, maybe my substitution is not the best approach. Let me try a different route.Going back to the equation:[b^2 - 2^{2n} = 2^n m^2]Let me set ( b = 2^{n} k + c ), where ( k ) and ( c ) are integers to be determined. Maybe this can help me express ( b ) in terms of ( 2^n ) and find suitable ( c ) and ( k ).But this might complicate things further. Alternatively, perhaps I can find a parametric solution for ( b ) and ( m ).Let me consider that ( b^2 - 2^{2n} = 2^n m^2 ). Let me divide both sides by ( 2^{2n} ):[left( frac{b}{2^n} right)^2 - 1 = left( frac{m}{2^{n/2}} right)^2]But ( m ) is an integer, so ( 2^{n/2} ) must divide ( m ). However, ( m ) is odd, so ( 2^{n/2} ) must be 1, which implies ( n = 0 ). Again, this only works for ( n = 0 ), which is not helpful for ( n geq 1 ).Wait, maybe I need to approach this differently. Let me think about specific cases to get an idea.**Case 1: ( n = 0 )**Then, the area should be ( (2^0 m)^2 = m^2 ), where ( m ) is odd.Choose ( a = 1 ), ( b = 0 ). Then the vertices are ( (1, 1) ), ( (-1, 1) ), and ( (0, 0) ).Calculating the area:[text{Area} = frac{1}{2} |1(1 - 0) + (-1)(0 - 1) + 0(1 - 1)| = frac{1}{2} |1 + 1| = 1]Which is ( 1^2 ), so ( m = 1 ), which is odd. Great, works for ( n = 0 ).**Case 2: ( n = 1 )**We need the area to be ( (2^1 m)^2 = (2m)^2 ), where ( m ) is odd.Let me choose ( a = 2 ). Then, the area is ( 2 |b^2 - 4| ). We need this to be ( (2m)^2 = 4m^2 ).So:[2 |b^2 - 4| = 4m^2][|b^2 - 4| = 2m^2]So, ( b^2 - 4 = pm 2m^2 ).Let me consider ( b^2 - 4 = 2m^2 ). Then:[b^2 = 2m^2 + 4]Looking for integer solutions. Let me try small values of ( m ):- ( m = 1 ): ( b^2 = 2 + 4 = 6 ), not a square.- ( m = 2 ): ( b^2 = 8 + 4 = 12 ), not a square.- ( m = 3 ): ( b^2 = 18 + 4 = 22 ), not a square.- ( m = 4 ): ( b^2 = 32 + 4 = 36 ), which is ( 6^2 ). So, ( b = 6 ).Thus, ( m = 4 ), but ( m ) needs to be odd. Hmm, that's a problem. Alternatively, maybe ( b^2 - 4 = -2m^2 ):[b^2 = 4 - 2m^2]But ( b^2 ) must be non-negative, so ( 4 - 2m^2 geq 0 ), which implies ( m^2 leq 2 ). So, ( m = 1 ):[b^2 = 4 - 2 = 2]Not a square. So, no solution here. Hmm, seems like for ( n = 1 ), choosing ( a = 2 ) doesn't work because ( m ) ends up even.Maybe I need to choose a different ( a ). Let me try ( a = 1 ). Then, the area is ( 1 |b^2 - 1| ). We need this to be ( (2m)^2 = 4m^2 ).So:[|b^2 - 1| = 4m^2]Which gives ( b^2 - 1 = pm 4m^2 ).First, ( b^2 - 1 = 4m^2 ):[b^2 = 4m^2 + 1]Looking for integer solutions. Let me try small ( m ):- ( m = 1 ): ( b^2 = 4 + 1 = 5 ), not a square.- ( m = 2 ): ( b^2 = 16 + 1 = 17 ), not a square.- ( m = 3 ): ( b^2 = 36 + 1 = 37 ), not a square.- ( m = 4 ): ( b^2 = 64 + 1 = 65 ), not a square.- ( m = 5 ): ( b^2 = 100 + 1 = 101 ), not a square.Not promising. Now, the other case: ( b^2 - 1 = -4m^2 ):[b^2 = 1 - 4m^2]Again, ( b^2 geq 0 ), so ( 1 - 4m^2 geq 0 ), which implies ( m^2 leq 1/4 ). Thus, ( m = 0 ), but ( m ) must be positive. So, no solution here either.Hmm, this is tricky. Maybe my initial choice of vertices is not the best. Instead of choosing ( (a, a^2) ), ( (-a, a^2) ), and ( (b, b^2) ), perhaps I need a different configuration.Alternatively, maybe I should consider three arbitrary points on the parabola and find a general expression for the area. Let me recall that the area can also be expressed using determinants.Given three points ( (x_1, y_1) ), ( (x_2, y_2) ), ( (x_3, y_3) ), the area is:[text{Area} = frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|]Plugging in ( y_i = x_i^2 ):[text{Area} = frac{1}{2} |x_1(x_2^2 - x_3^2) + x_2(x_3^2 - x_1^2) + x_3(x_1^2 - x_2^2)|]Factor the differences of squares:[= frac{1}{2} |x_1(x_2 - x_3)(x_2 + x_3) + x_2(x_3 - x_1)(x_3 + x_1) + x_3(x_1 - x_2)(x_1 + x_2)|]This still looks complicated. Maybe I can find a symmetric configuration where two points are symmetric about the y-axis, as I did before, but with a different third point.Wait, another idea: maybe use three points where one is at the vertex of the parabola, say ( (0, 0) ), and the other two are symmetric about the y-axis, say ( (a, a^2) ) and ( (-a, a^2) ). Then, the area would be:[text{Area} = frac{1}{2} |0(a^2 - a^2) + a(a^2 - 0) + (-a)(0 - a^2)| = frac{1}{2} |0 + a^3 + a^3| = frac{1}{2} times 2a^3 = a^3]So, the area is ( a^3 ). We need this to be ( (2^n m)^2 ). Thus:[a^3 = (2^n m)^2]Which implies:[a = (2^n m)^{2/3}]But ( a ) must be an integer, so ( (2^n m)^{2/3} ) must be integer. This requires that ( 2^n m ) is a perfect cube. Let me denote ( 2^n m = k^3 ), where ( k ) is an integer. Then:[a = k^2]So, ( m = frac{k^3}{2^n} ). Since ( m ) must be an integer, ( 2^n ) must divide ( k^3 ). Let me write ( k = 2^t s ), where ( s ) is odd. Then:[k^3 = 2^{3t} s^3][m = frac{2^{3t} s^3}{2^n} = 2^{3t - n} s^3]For ( m ) to be an integer, ( 3t - n geq 0 ), so ( t geq lceil n/3 rceil ). Also, ( m ) must be odd, which requires that ( 2^{3t - n} ) is 1, meaning ( 3t - n = 0 ), so ( n = 3t ). Therefore, ( n ) must be a multiple of 3.But the problem states that this must hold for every nonnegative integer ( n ), not just multiples of 3. So, this approach only works when ( n ) is a multiple of 3, which is insufficient.Hmm, so maybe choosing the vertex at ( (0, 0) ) is too restrictive. Let me go back to the previous configuration with two symmetric points and a third point.We had:[text{Area} = a |b^2 - a^2|]And we need this to be ( (2^n m)^2 ). So:[a |b^2 - a^2| = (2^n m)^2]Let me choose ( a = 2^{2n} ). Then:[2^{2n} |b^2 - 2^{4n}| = 2^{2n} m^2]Divide both sides by ( 2^{2n} ):[|b^2 - 2^{4n}| = m^2]So, ( b^2 - 2^{4n} = pm m^2 ). Let's take the positive case:[b^2 - m^2 = 2^{4n}][(b - m)(b + m) = 2^{4n}]Since ( b ) and ( m ) are integers, ( b - m ) and ( b + m ) are both powers of 2. Let me denote:[b - m = 2^k][b + m = 2^l]Where ( k < l ) and ( k + l = 4n ). Adding these two equations:[2b = 2^k + 2^l][b = 2^{k - 1} + 2^{l - 1}]Subtracting them:[2m = 2^l - 2^k][m = 2^{l - 1} - 2^{k - 1}]Since ( m ) must be odd, ( 2^{l - 1} - 2^{k - 1} ) must be odd. Let's analyze this expression.If ( k = 1 ), then:[m = 2^{l - 1} - 1]To make ( m ) odd, ( 2^{l - 1} ) must be even, which it is for ( l > 1 ). So, ( m ) will be odd as long as ( l > 1 ).Given that ( k + l = 4n ) and ( k = 1 ), then ( l = 4n - 1 ). Therefore:[b = 2^{0} + 2^{4n - 2} = 1 + 2^{4n - 2}][m = 2^{4n - 2} - 1]Since ( 4n - 2 geq 2 ) for ( n geq 1 ), ( m ) is indeed odd.Let me verify this for ( n = 1 ):- ( a = 2^{2 times 1} = 4 )- ( b = 1 + 2^{4 times 1 - 2} = 1 + 2^2 = 5 )- ( m = 2^{4 times 1 - 2} - 1 = 2^2 - 1 = 3 ) (which is odd)Calculating the area:[text{Area} = 4 |5^2 - 4^2| = 4 |25 - 16| = 4 times 9 = 36][(2^1 m)^2 = (2 times 3)^2 = 6^2 = 36]Perfect, it works for ( n = 1 ).Let me check ( n = 2 ):- ( a = 2^{4} = 16 )- ( b = 1 + 2^{8 - 2} = 1 + 2^6 = 65 )- ( m = 2^{6} - 1 = 64 - 1 = 63 ) (which is odd)Calculating the area:[text{Area} = 16 |65^2 - 16^2| = 16 |4225 - 256| = 16 times 3969 = 63504][(2^2 m)^2 = (4 times 63)^2 = 252^2 = 63504]Works as well.So, it seems that by choosing ( a = 2^{2n} ), ( b = 1 + 2^{4n - 2} ), and ( m = 2^{4n - 2} - 1 ), we satisfy the conditions for any nonnegative integer ( n ). For ( n = 0 ), we can choose ( a = 1 ), ( b = 0 ), and ( m = 1 ) as we did earlier.Therefore, this construction works for all ( n geq 0 ), providing the required odd ( m ) and the corresponding parabolic triangle with integer coordinates and the desired area.**Final Answer**boxed{(2^n m)^2}