Middle School Math Olympiad | Prove at least one of a and b is divisible by 3 | Pythagorean Triples

For Pythagorean Triples, how to prove for at least one of a and b is divisible by 3? We use the method of proof by contradiction to solve this problem. Basic number theory results are used to show that for a Pythagorean triple, at least one of a and b is divisible by 3. This is a Middle School Mathematical Olympiad Question.

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[valeriereid]

Thanks Dr. Wang this most certainly helped.

[haradhandatta]

Thanks, SIR. Nice proof by using modular arithmetic.
This can be proved by using elementary algebra.
Proof: Let us suppose that, none of a, b & c be divisible by 3.
Now, a^2 + b^2= c^2,
or, . a^2=c^2 -- b^2
=(c^2-1) - (b^2-1)
=(c-1)(c+1)-(b-1)(b+1)
But (c-1), c, (c+1) are three consecutive natural numbers.
So, any one of the factors must be divisible by 3 and c is not divisible by 3. So, either (c-1), or (c+1) is divisible by 3. We can take, (c-1)(c+1)=3
m, for some natural number m.
Similarly, (b-1)(b+1)=3n, for sy natural number n.
Hence, a^2=3m-3n=3(m-n).
So, a^2. & hence a. is divisible by 3.
Thus, . at least one of the Pythagorean triple is divisible by 3.

[faq]

Thank you SO MUCH. I am going through a proofs book and this one appeared, but the book didn't introduce proofs by contradiction yet;
After watching your (amazing) proof, and the other one in the comment section, I assume it is impossible to prove this without contradiction, because this is already pretty complicated for me, but I will ask anyways because of curiosity:
Is it possible to prove this without using contradiction?