If a^2 + b^ 2 + c^2 = abc...?

I know how to prove this. I just think the proof is interesting. If math is your forte, see if you can work it out...





Given three positive integers a, b, and c such that a^2 + b^2 + c^2 = abc. The number of solutions is infinite; for example, a = 3, b = 6, c = 15 is one. However, prove that for every solution, a, b, and c must all be divisible by 3.

If a^2 + b^ 2 + c^2 = abc...?
I see a way to solve this by breaking it up into cases and looking at the equation modulo 3. Suppose a,b, and c are not all divisible by three. Note that a number d not divisible by 3 must satisfy d^2 = 1 mod 3. There are two cases:





1) All three a,b, and c are not divisible by three. Then when you reduce mod 3 you get:





a^2 +b^2 +c^2 = 1+1+1 = 0 mod 3. However, abc is not 0 mod 3 since none of a,b, or c is divisible by 3.





2) At least one of a,b, or c is divisible by 3. Then once again reducing mod 3:





a^2 + b^2 + c^2 = either 1 or 2 mod 3. However abc = 0 mod 3 since one of the three is divisible by 3.





Thus both of the above cases lead to a contradiction, and so all of a, b, and c must be divisible by three.
Reply:Well, look at the squares mod 3. 0 and 1 are possible squares; 2 is not.





Given that, let us examine each of the possible cases for what the LHS is congruent to mod 3.





If it's congruent to 1, it is the sum of a 1 and 2 0s. But this means the RHS must be congruent to 0 mod 3. Contradiction.





A similar argument shows it cannot be congruent to 2. So it is congruent to 0. So the RHS is congruent to 0 too. So at least one of a, b, or c is divisible by 3; i.e., at least one of the terms on the LHS is congruent to 0. But the only way this can happen is if they all are.





QED.