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I suppose I need to use prime factorization. I want to show $p^2-q^2=24k$ for some integer $k$ . How can I start this proof?

Andrés E. Caicedo
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Wes
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3 Answers3

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Hint: Show $p^2 = 1 \mod 3$ and $p^2 = 1 \mod 8$ for any such $p$.

Zarrax
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  • You mean like show $p^2-1=3k$ , $k$ integer, and $p^2-1=8m$, $m$ is an integer? – Wes Sep 28 '13 at 01:42
  • But I don't have anything about $p$ except it is bigger than 5 and q.. I am so stuck – Wes Sep 28 '13 at 01:43
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    Use CRT to glue this two together, and you'll get $p^2 \equiv 1 \pmod {24}$, this holds for every prime number, so it also holds for $q^2$, so we have:

    $$p^2 - q^2 \equiv 1 - 1 \equiv 0 \pmod {24} \implies p^2 - q^2 = 24k $$

     – Stefan4024 Sep 28 '13 at 01:50
  • @Wes Yes.. you know other things, that $p$ is odd and that $p = 1$ or $-1 \mod 3$. What does that mean in terms of $p^2$? – Zarrax Sep 28 '13 at 01:59
  • But $p$ is bigger than 5? How can it be 1? – Wes Sep 28 '13 at 04:23
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    $p\color{red}\equiv\pm1\pmod3$ – J. W. Tanner Sep 08 '19 at 04:23
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Like Stefan,

$$(6a\pm1)^2=36a^2\pm12a+1=24a^2+24\frac{a(a\pm1)}2+1\equiv1\pmod{24}$$ as the product of two consecutive integers is always even

Observe that $6a\pm1$ is not necessarily prime, but $(6a\pm1,6)=1$

So, any number $p$ relatively prime to $6$ will satisfy this

lab bhattacharjee
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It's well-known fact that every prime number, except for $2$ and $3$ can be written as $6k \pm 1$. SO we have:

$$(6k \pm 1)^2 - (6l \pm 1)^2 = 36k^2 \pm 12k + 1 - 36l^2 \mp 12l - 1$$

$$= 36(k^2-l^2) \pm 12(k - l) = 12(3k^2 - 3l^2 \pm k \mp l)$$

The sum in the parenthesis is always even. Why?

Hence we proved that it's divisible by 24.

Stefan4024
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