# Perfect Square Arising from Prime Pythagorean Triple

This problem appears as the second question in the British Mathematical Olympiad 2014--2015 Round 1 paper (https://bmos.ukmt.org.uk/home/bmo1-2015.pdf).

Positive integers $p$, $a$ and $b$ satisfy the equation $p^2 + a^2 = b^2$. Prove that if $p$ is a prime greater than $3$, then $a$ is a multiple of $12$ and $2(p + a + 1)$ is a perfect square.

Using the difference of two squares, unique prime factorisation theorem and properties of the product of subsequent integers, I have managed to prove that $a$ is a multiple of 12 (i.e. that $a = 12q$ for $q \in \mathbb{N}$).

However, for the second part, I am not sure how to link the Pythagorean theorem to the required result, given that $a$ is a multiple of $12$. Since $2(p + a + 1)$ is even, then it is easy to deduce that we are looking for an expression for $t$ satisfying $p + a + 1 = 2t^2$.

Any clues/hints on tackling this problem would be greatly appreciated.

## 2 Answers

We have $p^2=b^2-a^2=(b-a)(b+a)$. Note that $p$ cannot divide both of $b-a$ and $b+a$. For if it did, then $p$ would divide both $a$ and $b$, and we would have $1^2+(a/p)^2=(b/p)^2$, which is impossible.

So we must have $b-a=1$ and $b+a=p^2$. It follows that $a=\frac{p^2-1}{2}$, and therefore $$2(p+a+1)=2p+p^2-1+2=(p+1)^2.$$

• Why p > 3? Works for 3-4-5 triple too. – Oscar Lanzi Mar 14 '16 at 0:27
• The $p\gt 3$ condition was for proving that $12$ divides $a$. Since OP mentioned having done that part of the problem, I did not deal with it. Certainly the second part does not require $p\gt 3$. – André Nicolas Mar 14 '16 at 0:30

All pythagorean triple is of the form $(m^2-n^2,2mn,m^2+n^2)$ and the prime $p$ must be equal to $m^2-n^2$ which force to take $m=n+1$ so $p=2n+1$ with $n>1$.

It follows $a=2n(n+1)$ so $a$ is clearly multiple of $4$.

Besides the number $n$ have three possible cases $n=3n_1$,$\space n=3n_1+1$ and $n=3n_1-1$ and the only that make not clear that $a$ is not multiple of $3$ is when $n=3n_1+1$. However for this case we would have $p=6n_1+3$ which is not admissible because $p>3$ and prime.

It is not problem to verify that $2(p+a+1)=(2(n+1))^2$