Problem 1650

Solution:

a.
Since $a$ and $b$ have the same parity, I can set $a = x + y$ and $b = x - y$ and the equation becomes $2 c^{2} = 2 x^{2} + 2 y^{2}$ and therefore the smallest $c$ is given by the smallest Pythagorean triple $(x, y, c) = (3, 4, 5)$, which gives $(a, b, c) = (7, 1, 5)$.

b.
It suffices to observe that $(7k, k, 5k)$ is also a solution for every positive $k$.

Solution:

a.
The equation is symmetric in $a$ and $b$, let us solve for $a > b$. It is convenient to set $a = c + n$ and $b = c - m$, $n, m > 0$. The equation then becomes $2c(m - n) = n^{2} + m^{2}$. Hence $2$ divides $n^{2} + m^{2}$, so $n$ and $m$ have the same parity. We can then perform a new change of variables $\alpha = \frac{n + m}{2}$ and $\beta = \frac{m - n}{2}$, with $\alpha > \beta$. We have $2c\beta = \alpha^{2} + \beta^{2}$. For the same reasons as before we obtain that $\alpha$ and $\beta$ have the same parity, moreover we must have $\beta \mid \alpha^{2} + \beta^{2}$, that is $\beta \mid \alpha$. Let us set $\alpha = k\beta$. We then have $c = \frac{(k^{2} + 1)\beta}{2}$. This is an increasing function both with respect to $k$ and with respect to $\beta$. Therefore I will have to take $k$ and $\beta$ as small as possible. The condition $\alpha > \beta$ imposes $k \geq 2$, while I have no conditions on $\beta$, so $\beta \geq 1$. If $\beta = 1$ then $k^{2} + 1$ must be even and therefore the smallest possible value of $k$ is $3$, if $\beta = 2$ I can also take $k = 2$. For larger values of $\beta$, the minimum $k$ that I can always take is $k = 3$ for odd $\beta$ and $k = 2$ for even $\beta$, and since $c$ is increasing in $\beta$, these values will turn out to be larger, respectively, than the one obtained with $(k, \beta) = (3, 1)$ for odd $\beta$ and than the one obtained for $(k, \beta) = (2, 2)$ for even $\beta$.

Computing the two values of $c$ for these two pairs one obtains $c = 5$ in both cases. We now need to see whether at least one of these two pairs $(k, \beta)$ gives a pair $(a, b)$ of positive integers. By reversing the steps we obtain $(a, b) = (7, 1)$ for the pair $(k, \beta) = (3, 1)$ and $(a, b) = (7, -1)$ for $(k, \beta) = (2, 2)$. Only the pair $(7, 1)$ is therefore acceptable. Besides this one we must also take, by symmetry, $(1, 7)$. The triples are therefore $(7, 1, 5)$ and $(1, 7, 5)$. And the minimum $c$ is $5$.

b.
It suffices to observe that for every odd $k$ with $k \geq 3$ I obtain a triple of integers $(a, b, c)$ that is a solution, indeed:
$$\text{k = 2h + 3 (a, b, c) = (2(h + 2) 2 - 1 ) , (2(h + 1) 2 - 1 ) , (2h 2 + 6h + 5 ) )}$$
with $a, b, c > 0$

Or that for even $k$ with $k \geq 4$ and for every even $\beta$ one obtains a triple of integers $(a, b, c)$ that is a solution, indeed
$$\text{k = 2h + 4 = 2 (a, b, c) = ((2h + 5) 2 - 2 ) , ((2h + 3) 2 - 2 ) , (4h 2 + 8h + 5 ) )}$$
with $a, b, c > 0$.