How do I prove that if $7\mid a^2+b^2$, then $7\mid a$ and $7\mid b$?
I am not allowed to use modular arithmetic.
Assuming $7$ divides $a^2+b^2$, how do I prove that the sum of the squares of the residuals is 0?
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parkhyeyoo
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Look at the quadratic residues – Rellek May 16 '15 at 19:55
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Actually, are you familiar with modular arithmetic? Or do you only know principles of divisibility? – Rellek May 16 '15 at 19:56
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No, I am not allowed to use modular arithmetic. – parkhyeyoo May 16 '15 at 19:59
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I would argue by contrapositive then. Try assuming that 7 divides neither a nor b, and conclude that there is no way it can divide the sum of their squares. – Rellek May 16 '15 at 20:00
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Rellek, I did that but how do I proof that if $7|(a^2+b^2)$ then the sum of the squares of the residuals is 0? – parkhyeyoo May 16 '15 at 20:11
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More generally: $\ p\equiv 3\pmod{\! 4},\ \, p\mid a^2+b^2\,\,\Rightarrow\,\, p\mid a,b$.
By contradiction: if not, then mod $p$: $\ a^2\equiv -b^2\iff (a/b)^2\equiv -1$
$1)\ \stackrel{(p-1)/2}\Rightarrow(a/b)^{p-1}\equiv \color{#00F}{{(-1)^{(p-1)/2}}}\equiv \color{#00F}{-1}\, $ contradicts little Fermat.
$2)\ \stackrel{2}\Rightarrow\, (a/b)^4\equiv 1\,$ and $\,\text{ord}_p (a/b)=4\mid p-1$,$\,$ contr $\,p\equiv 3\,\,$ mod $4$.
It is a consequence of $\,p\nmid a,b,\,\,p\mid a^2+b^2\,\Rightarrow\, p\equiv 1\,\,$ mod $4\,$ (see here).
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Talk about terse! Correct me if I'm wrong, but you are presenting two independent proofs here: in $(1)$, you raise both sides to the power $(p-1)/2$, and in $(2)$, you square both sides. (Or is $$\stackrel{(p-1)/2}\Rightarrow$$ a standard notation that I'm too old-fashioned to understand?) – TonyK Jun 18 '15 at 10:17
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There is a theorem in number theory which asserts
If $p|a^2+b^2$, and $p\equiv 3\pmod4$, then $p|a,p|b$.
Proof. Suppose not,then by $p|a^2+b^2$, we have $a^2\equiv -b^2\pmod p$, and by using Fermat's little theorem we have $$ 1\equiv a^{p-1}\equiv (a^2)^{(\frac{p-1}2)}\equiv (-b^2)^\frac{p-1}2\equiv(-1)^{\frac{p-1}2}=-1\pmod p $$ Which contradicts...
Now observe that $7\equiv3\pmod4$
k1.M
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Since you said that you are not allowed to use modular arithmetic, argue by the contrapositive. If 7 does not divide $a$ or $b$. Then,
$a = 7k \pm 1, 7k \pm 2, 7k \pm 3 \implies$
$a^2 = 7(7k^2 \pm 2k) + 1, 7(7k^2 \pm 4k) +4, 7(7k^2 +8k + 1)+2$, $k \in \Bbb Z$
Now, look at any linear combination of $a^2$ (with integer weights, or course) and it is easy to convince oneself that 7 will never divide the sum.
Rellek
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