High Five, Pythagoras!

Relevant wiki: Pythagorean Triples

A square is either 0, 1, or 4 modulo 5. If both $a$ and $b$ are not multiples of 5, then the only possible cases are as follows: $\begin{array}{c|c|c} a^2 \pmod{5} & b^2 \pmod{5} & a^2 + b^2 \pmod{5} \\ \hline 1 & 1 & 2 \\ 1 & 4 & 0 \\ 4 & 1 & 0 \\ 4 & 4 & 3 \end{array}$ Since $a^2 + b^2 = c^2$ is a square, we can rule out the first case and fourth case. In the only cases left, $c$ is divisible by 5.

Hence, at least one of $a$ , $b$ , $c$ is divisible by 5.

In fact, is also true that one of them must be divisible by 3, and divisible by 4. I'll leave this as an exercise.

Relevant wiki: Pythagorean Triples

A basic fact about Pythagorean triples is, that they may be generated by the equations $\begin{cases} a = p^2 - q^2 \\ b = 2pq \\ c = p^2 + q^2 \end{cases}$ where $p > q$ are positive integers, and $a$ and $b$ may need to be swapped.

Now for any integer $x$ , $x^2 \equiv -1, 0, 1$ mod 5. Therefore consider the following possibilities.

• At least one of $p, q \equiv 0$ mod 5. Then $b = 2pq \equiv 0$ mod 5.

• $p^2 \equiv q^2 \equiv \pm 1$ mod 5. Then $a = p^2 - q^2 \equiv (\pm 1) - (\pm 1) \equiv 0$ mod 5.

• $p^2 \equiv -q^2 \equiv \pm 1$ mod 5. Then $c = p^2 + q^2 \equiv (\pm 1) + (\mp 1) \equiv 0$ mod 5.

In each of these cases, at least one of the numbers in the triple is a multiple of five. Therefore there no triples that match the description; the answer is $\boxed{0}$ .

Let $a = m^2 - n^2, b =2mn, c = m^2 + n^2$ be a primitive Pythagorean triple. Let's prove that at least one of $a,b,c$ must be divisible by 5. Fortunately for us, 5 is prime so if one of them is divisible by 5 then the product $abc$ will be divisible by 5 (and vice-versa). So let's compute $abc = 2m^5n - 2mn^5$ and by applying Fermat's little theorem we see that $abc \equiv 2mn - 2mn \equiv 0 \mod 5$ . Therefore there are 0 of these Pythagorean triples.

Perfect squares can only have a last digit of 0, 1, 4, 5, 6, or 9. Excluding multiples of 5, they must end in 1, 4, 6, or 9. Adding two such numbers together to get $c^{2}$ , we find that no such sum gives us a number ending in 1, 4, 6, or 9. So, the answer is zero.

But, I prefer Jon Haussmann's solution. Mine is closely related to his, but not as elegant in my opinion.

By logic, we know that there is no way to know for sure while doing trial and error because 1000 is a very large number. So the smart thing to do would be to answer 0 because after several attempts, a pattern will emerge where in some way or the other, there is always a multiple of 5.