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.


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 $$2(p+a+1)=2p+p^2-1+2=(p+1)^2.$$

  • $\begingroup$ Why p > 3? Works for 3-4-5 triple too. $\endgroup$ – Oscar Lanzi Mar 14 '16 at 0:27
  • 2
    $\begingroup$ 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 $p\gt 3$. $\endgroup$ – 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 $2(p+a+1)=(2(n+1))^2$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.