For m,n,p belong to Z, suppose that 5 divides m^2+n^2+p^2. Prove that 5 divides at least one of {m,n,p}. (i.e. Prove if m^2+n^2+p^2 is congruent to 0 mod 5, then m is congruent to 0 mod 5, n is congruent to 0 mod 5 or p is congruent to 0 mod 5.)
For m,n,p belong to Z, suppose that 5 divides m^2+n^2+p^2. Prove that 5 divides at least one of {m,n,p}. (i.e. Prove if m^2+n^2+p^2 is congruent to 0 mod 5, then m is congruent to 0 mod 5, n is congruent to 0 mod 5 or p is congruent to 0 mod 5.)
I don't really get how this would help with the proof.
I did like this:
Assume none of {m,n,p} is divisible by 5. This implies that none of {m^2,n^2,p^2} is divisible by 5, i.e.
m^2 ≢ 0 mod 5
n^2 ≢ 0 mod 5
p^2 ≢ 0 mod 5
m^2 + n^2 + p^2 ≢ 0 mod 5 which leads to a contradiction => one of {m,n,p} must be divisible by 5.
Is this a valid proof?
My browser does not show any symbol between "m^2" and "0 mod 5". I assume you mean m^2 is not congruent to 0 modulo 5. It is not clear why you conclude that m^2 + n^2 + p^2 is not congruent to 0. For example, if m^2 = 3 and n^2 = p^2 = 1 mod 5, then m^2 + n^2 + p^2 = 0 mod 5. The problem is that m^2 can't be 3 mod 5.
I also suggested proving this fact by contradiction and finding all possible values of m^2 mod 5. Then see if three such values can add up to zero.
No. Maybe $\displaystyle m^2 \equiv n^2 \equiv 2$ and $\displaystyle p^2 \equiv 1$ (mod 5).
Suppose m,n,p are not 0 (mod 5). Modulo 5
$\displaystyle 1^2 \equiv 1$
$\displaystyle 2^2 \equiv 4$
$\displaystyle 3^2 \equiv 4$
$\displaystyle 4^2 \equiv 1$
It is impossible to have three such squares add up to 0 (mod 5). Therefore at least one of m,n,p is congruent to 0 (mod 5).