15. ILLUSTRATIVE EXERCISE

a) What is the smallest positive integer $c$ such that there exists at least one pair $(a, b)$ of distinct positive integers such that $2 c^{2}=a^{2}+b^{2}$?

b) Prove that there are infinitely many triples $(a, b, c)$ of distinct positive integers such that $2 c^{2}=a^{2}+b^{2}$. SOLUTION
15. Question 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 thus the smallest $c$ is given by the smallest Pythagorean triple $(x, y, c)=(3,4,5)$, which gives $(a, b, c)=(7,1,5)$.

Question b)

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

Second Solution

Question a)

The equation is symmetric in $a$ and $b$, let's solve for $a>b$. It is convenient to set $a=c+n$ and $b=c-m$, with $n, m>0$. The equation then becomes $2 c(m-n)=n^{2}+m^{2}$. Therefore, 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 $2 c \beta=\alpha^{2}+\beta^{2}$. For the same reasons as before, we get that $\alpha$ and $\beta$ have the same parity, and moreover, we must have $\beta \mid \alpha^{2}+\beta^{2}$, i.e., $\beta \mid \alpha$. Let $\alpha=k \beta$. Then we have $c=\frac{\left(k^{2}+1\right) \beta}{2}$. This is an increasing function both with respect to $k$ and with respect to $\beta$. Therefore, I will have to take the smallest possible $k$ and $\beta$. The condition $\alpha>\beta$ imposes $k \geq 2$, while there are no conditions on $\beta$, so $\beta \geq 1$. If $\beta=1$, then $k^{2}+1$ must be even, and thus the smallest possible value of $k$ is 3. If $\beta=2$, I can also take $k=2$. For larger values of $\beta$, the smallest $k$ 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 be greater than those obtained with $(k, \beta)=(3,1)$ for odd $\beta$ and with $(k, \beta)=(2,2)$ for even $\beta$.

Calculating the two values of $c$ for these two pairs, we get $c=5$ in both cases. We now need to see if at least one of these two pairs $(k, \beta)$ gives a pair $(a, b)$ of positive integers. Reversing the steps, we get $(a, b)=(7,1)$ 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. In addition to this, we must take for symmetry also $(1,7)$. The triples are therefore $(7,1,5)$ and $(1,7,5)$. And the minimum $c$ is 5.

Question b)

It is enough to observe that for every odd $k$ and $k \geq 3$ I get a triple of $(a, b, c)$ integers that is a solution, indeed:

$$
\begin{gathered}
k=2 h+3 \\
(a, b, c)=\left(2(h+2)^{2}-1\right) \beta,\left(2(h+1)^{2}-1\right) \beta,\left(2 h^{2}+6 h+5\right) \beta)
\end{gathered}
$$

with $a, b, c>0$

Or that for even $k$ and $k \geq 4$ and for every even $\beta$ I get a triple of integers $(a, b, c)$ that is a solution, indeed:

$$
\begin{gathered}
k=2 h+4 \\
\beta=2 \gamma \\
(a, b, c)=\left(\left((2 h+5)^{2}-2\right) \gamma,\left((2 h+3)^{2}-2\right) \gamma,\left(4 h^{2}+8 h+5\right) \gamma\right)
\end{gathered}
$$

with $a, b, c>0$.