1
$\begingroup$

I'm reading a proof in my elementary number theory textbook that no positive integer of the form $4^n (8m+7)$ can be represented as the sum of three squares. I'm caught up on one detail of the proof, where they say:

Let us suppose that $4^n (8m+7)$, where $n \geq 1$, can be written as $$ 4^n (8m+7) = a^2 + b^2 + c^2. $$ Then each of the integers $a,b,c$ must be even.

How do they conclude that $a,b,c$ are all even? Why couldn't two of them be odd, and the other one be even? In that case, the sum of their squares will be even, but then how can one show it wouldn't be a multiple of $4$?

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ The square of an odd number, modulo $4$, is always $1$ and never $3$. $\endgroup$ – Rahul Dec 11 '13 at 1:50
  • 1
    $\begingroup$ The square of an odd number is $\equiv 1 \pmod{4}$, so if two of the numbers were odd and the third even, the sum would be $\equiv 2 \pmod{4}$. $\endgroup$ – Daniel Fischer Dec 11 '13 at 1:50
1
$\begingroup$

Hint: $(2n+1)^2=4n(n+1)+1$. ...

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.