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Here is what I have so far:

Because 7 divides a^2+b^2, this means that a^2+b^2 ≡ 0 (mod 7). From this, we deduce that a^2 ≡ 0 (mod 7) and b^2 ≡ 0 (mod 7). For a^2 ≡ 0 (mod 7), we multiply both sides by a^(-1) to get a ≡ 0 (mod 7). This means that a is divisible by 7. Likewise, For b^2 ≡ 0 (mod 7), we multiply both sides by b^(-1) to get b ≡ 0 (mod 7), which implies that b is divisible by 7. Hence, both a and b are divisible by 7.

I think this is the correct way to prove it but I just want to make sure that my steps are correct and that I'm not missing anything.

FoiledIt24
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    You can't conclude that $a^2 \equiv 0$ just because $a^2 + b^2 \equiv 0$. Arguments that rely on the positivity of squares make no sense in modular arithmetic. Hint: What are all the possible values of $a^2$ modulo $7$? What can you get by adding two of them? (in the future, please use mathjax when asking here. https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Ethan Bolker Apr 06 '19 at 02:55
  • Would that still work if you replaced $7$ by $13$? – Angina Seng Apr 06 '19 at 02:56
  • Special case of this. and many others. – Bill Dubuque Apr 06 '19 at 03:15

1 Answers1

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Suppose neither a nor b were divisible by 7, then the options are

Congruent to 1, square is congruent to 1 Congruent to 2, square is congruent to 4 Congruent to 3, square is congruent to 2 Congruent to 4, square is congruent to 2 Congruent to 5, square is congruent to 4 Congruent to 6, square is congruent to 1

Take any pair of those 1,2 or 4. The possibilities for the sum of those are 2,3,5,6,1 for congruence mod 7. Notice that you can't get 0.

So by contradiction, at least one of a and b must be divisible by 7.

AHusain
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