Positively hilarious

Using the quadratic formula, the problem essentially asks the following: For what $(b, c)$ are there such that both $b^2 + 4c$ and $b^2 - 4c$ are both perfect squares. We know that that condition is sufficient to be an integer because the numerator will always be even, thus, the whole expression will be an integer.

If the difference between two squares is even, then it is a multiple of $4$ . This can be shown with the following:

The parity of $b^2 + 4c = m^2$ and $b^2 - 4c = n^2$ are the same, thus the parity of $m$ and $n$ are the same and they have an even difference. Rewriting $n$ as $(m + 2k)$ , we get $m^2 - (m + 2k)^2 = 4p \rightarrow -4m - 4 = 4p$ . So we can focus only on $b$ , restating the problem as "what integers satisfy the equation $x^2 + y^2 = 2b^2$ ?"

The only solutions for $p$ to the equation $m^2 + n^2 = p^2$ are primes of the form $4k + 1$ and multiples thereof. It follows that $(m+n)^2 + (m-n)^2 = 2p^2$ , thus $b$ is a prime of the form $4k + 1$ and multiples thereof.

Thus, the solutions for $b \leq 50$ are the multiples of $5, 13, 17, 29, 37, 41$ less than $50$ , which when added together equals $511$ .

The quadratic $x^2 + bx + c$ has integer roots if and only if its discriminant $b^2 - 4c$ is a perfect square. (If you are really interested in the pesky details, I have saved the justification for this somewhat obvious claim for the end of the proof.) Similarly, the quadratic $x^2 + bx - c$ has integer roots if and only if $b^2 + 4c$ is a perfect square.

Let $b^2 - 4c = x^2$ and $b^2 + 4c = y^2$ , where $x$ and $y$ are nonnegative integers. Since $b^2 + 4c$ is positive, $y$ is also positive. If $x = 0$ , then $b^2 = 4c$ , so $y^2 = b^2 + 4c = 2b^2$ , which is impossible (since $\sqrt{2}$ is irrational), so $x$ is also a positive integer. Also, $b^2 + 4c > b^2 - 4c$ , so $y > x$ .

Adding the equations $b^2 - 4c = x^2$ and $b^2 + 4c = y^2$ , we get $2b^2 = x^2 + y^2,$ so $b^2 = \frac{x^2 + y^2}{2}.$ Note that we can write this equation in the form $b^2 = \left( \frac{y + x}{2} \right)^2 + \left( \frac{y - x}{2} \right)^2.$

From the equations $b^2 - 4c = x^2$ and $b^2 + 4c = y^2$ , we see that $b$ , $x$ , and $y$ must all have the same parity. Hence, $(y + x)/2$ and $(y - x)/2$ are positive integers.

Let $u = (y + x)/2$ and $v = (y - x)/2$ , so $b^2 = u^2 + v^2.$ This equation tells us that $b$ is the largest number in a Pythagorean triple.

Conversely, suppose that $b$ is the largest number in a Pythagorean triple, so $b^2 = u^2 + v^2$ for some positive integers $u$ and $v$ . If both $u$ and $v$ are odd, then $u^2 \equiv v^2 \equiv 1 \pmod{4}$ , and $b^2 = u^2 + v^2 \equiv 2 \pmod{4},$ but no square is congruent to 2 modulo 4, so at least one of $u$ and $v$ must be even. Let $c = uv/2$ , which is a positive integer. Then $b^2 - 4c = u^2 + v^2 - 2uv = (u - v)^2,$ and $b^2 + 4c = u^2 + v^2 + 2uv = (u + v)^2,$ so both $x^2 + bx + c$ and $x^2 + bx - c$ have integer roots.

The problem now is to determine how many $b \le 50$ are the largest number in a Pythagorean triple. Recall that in a primitive Pythagorean triple, the largest number is of the form $m^2 + n^2,$ where $m$ and $n$ are relatively prime and have opposite parity. By hand, we can determine that the numbers of these form up to 50 are 5, 13, 17, 25, 29, 37, and 41. We can then take all multiples of these numbers that are at most 50. This gives us the list 5, 10, 13, 15, 17, 20, 25, 26, 29, 30, 34, 35, 37, 39, 40, 41, 45, and 50. The sum of these numbers is 511.

Claim . The quadratic $x^2 + bx + c$ has integer roots if and only if $b^2 - 4c$ is a perfect square.

Proof . Assume that the quadratic $x^2 + bx + c$ has integer roots. Then by the quadratic formula, $b^2 - 4c$ must be the square of a rational number. But $b^2 - 4c$ is an integer, so if it is the square of a rational number, then it must be a perfect square.

Conversely, assume that $b^2 - 4c$ is a perfect square. Then by the quadratic formula, $x^2 + bx + c$ has rational roots. But the quadratic $x^2 + bx + c$ is monic, so by the Rational Root theorem, its roots are integers.