Hey zenith20.
The formal way of stating this would be to prove that for:
a^2 + b^2 = c^2 then a = 5x (x is an integer) and gcd(a,b) = gcd(a,c) = 1 since 5 is a prime number.
is a pythogorean triple with more than one multiple of 5. Hence I will assume you are talking about primitive pythogorean triples.
We will prove your claim by contradiction. By definition, the primitive triples should not share a common factor. Suppose two of them shared a common factor of 5, the Pythogorus theorem would then imply that the third one should also contain a factor of 5. Then all of them would have a common factor of 5 contradicting primitiveness of the triple.
That can be answered by simple modular arithmetic:
Consider a pythogorean triple such that . Let us supposed that neither x nor y has a factor of 5. Then observe that can leave a remainder of 1 or 4 when divided by 5.
If , then . But a square of a number cannot be 2 mod 5.
If , then . But a square of a number cannot be 3 mod 5.
Thus one of the must be 1 mod 5 and the other one must be 4 mod 5. Adding we get that z is divisible by 5.
So we have proved that if x and y both are not divisible by 5 then z is divisible by 5.