I want to prove if following is true for every integer a,b and c
$$a^2 - b^2 = cp $$
then p|(a+b) or p|(a-b) where p is a prime number. Any suggestion, help would be highly appreciated. Thanks in advance
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2Since $a^2-b^2=(a+b)(a-b)$, you have $cp = (a+b)(a-b)$, hence $p | (a+b)(a-b)$, hence $p|(a+b)$ or $p|(a-b)$. – mjqxxxx Nov 13 '13 at 17:13
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how can you conlude just like that? – Bledi Boss Nov 13 '13 at 17:30
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One can either define prime numbers by that property, or appeal to unique prime factorization. If $p \mid xy$, then $p$ is in the factorization of $xy$, so it's either a factor of $x$ or of $y$. – Henry Swanson Nov 13 '13 at 19:50
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$$a^2-b^2=cp\implies a^2\equiv b^2\pmod p$$
If $p|a, b^2\equiv0\pmod p\implies b\equiv0$
Else $(ab,p)=1\implies \left(\frac ab\right)^2\equiv1\pmod p$
Now use this
lab bhattacharjee
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@BlediBoss, from the link, we have $\frac ab\equiv\pm1\pmod p\implies a\equiv\pm b\pmod p$ – lab bhattacharjee Nov 13 '13 at 17:45
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@BlediBoss, or as mjqxxxx has pointed out $\frac{(a+b)(a-b)}p=c$ which is an integer $\implies p$ must divide at least one of $a+b,a-b$ – lab bhattacharjee Nov 13 '13 at 17:46
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I mean before we come to that step. My aim was to show p|(a+b) or p|(a-b) . And one more thing, I guess it should be gcd(ab,p) = 1 at your answer. – Bledi Boss Nov 13 '13 at 17:48
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@BlediBoss, the question does not say $(ab,p)=1$ or not? From the link, we know $x^2\equiv1\pmod p\iff x\equiv\pm1$ – lab bhattacharjee Nov 13 '13 at 17:50
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yeah that's true. Once more you mean (ab,p)=1 or gcd(ab,p) = 1 ? And what has that to do with the answer p|(a+b) or p|(a-b)? I'm so confused – Bledi Boss Nov 13 '13 at 17:57
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