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Show that for any odd $n$ it follows that $n^2 \equiv 1 \mod 8$ and for uneven primes $p\neq 3$ we have $p^2 \equiv 1\mod 24$.

My workings so far: I proceeded by induction. Obviously $1^2 \equiv 1 \mod 8$. Then assume that for a certain uneven number $k$ we have $k^2 \equiv 1 \mod 8$. Then the the next uneven number is $k+2$ and $(k+2)^2 = k^2 + 4k + 4 \equiv 4k + 5 \mod 8$. by the induction hypothesis. Now since $k$ is uneven we can write it as $k=2j+1$ and thus $4k+5 =8j+9 \equiv 1 \mod 8$ and we have shown what was asked for by induction.

However, in the case of a prime number $n$ I am not so certain how to proceed because I can't use induction. It comes down to proving that $24|(p^2-1)$. This is certainly the case for the first uneven prime $p \neq 3$, namely $p=5$ such that $p^2-1 =24$. How would I proceed from here, or how should I approach the problem differently? Many thanks in advance!

wythagoras
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Slugger
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    In English, uneven is usually called odd. – Sammy Black May 03 '13 at 22:31
  • You almost have it. Certainly $p$ is odd (uneven), so $8$ divides $p^2-1$. Now, do you see why it suffices to prove that $3$ divides $p^2-1$? – Andrés E. Caicedo May 03 '13 at 22:32
  • @Andres Caicedo Yeah I understand the intuition that. $p^2-1$ can only be a multiple of both 3 and 8 if it is a multiple of $3\cdot 8$ because their greatest common denominator is 1. I don't really know how to formalize this though. – Slugger May 03 '13 at 22:42
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    The key to show that if $a,b$ are relatively prime, and both $a$ and $b$ divide $n$, then also $a\times b$ divides $n$ is that there is a combination $ax+by$ with integer coefficients that equals $1$. In this case, $1=9-8=3\times 3-8\times 1$. We can use this as follows: Say $n=3a=8b$. Then $a=1\times a=(9-8)a=3(3a)-8a=3(8b)-8a=8(3b-a)$ is a multiple of $8$, say $a=8t$ ($t$ is actually $3b-a$, but never mind that), so $n=3a=3\times 8t=24t$. – Andrés E. Caicedo May 03 '13 at 22:58
  • Cool, thanks a lot for your explanation! – Slugger May 03 '13 at 23:01

3 Answers3

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If $n$ is odd, $n = 2 k + 1$ and $n^2 = 4 k^2 + 4 k + 1 = 4 k (k + 1) + 1$, and $4 k (k + 1)$ is divisible by 8. Thus $n^2 \equiv 1 \pmod{8}$.

If $p > 3$ is prime, it is odd, and so by the above $p^2 \equiv 1 \pmod{8}$. As it is prime, $p \equiv 1 \pmod{3}$ or $p \equiv 2 \pmod{3}$, so $p^2 \equiv 1 \pmod{3}$. Now, as $\gcd(3, 8) = 1$, combining the above congruences gives $p^2 \equiv 1 \pmod{24}$.

vonbrand
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Never use induction at all. It is far too heavy a weapon for this kind of question.

  1. An odd number is a number of the form $4n±1$. Square it, and you get $16n^2±8n+1$, which clearly equals 1 modulo 8.
  2. Primes are irrelevant: this is true for all odd numbers which are not multiples of 3. Such a number is of the form $3n±1$. Square it, and you get $p^2=9n^2±6n+1$, which clearly equals 1 modulo 3. If this doesn't make it obvious to you that $p^2$ is 1 modulo 24, consider $p^2-1$. It is a multiple of 8, and it is also a multiple of 3, so it must be a multiple of 24.
Martin Kochanski
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For the second question, any prime $>3,$ can be expressed as $6a\pm1$ where $a$ is a positive integer

So, $p^2=(6a\pm1)^2=36a^2\pm12a+1=24a^2+24\cdot\frac{a(a\pm1)}2+1\equiv1\pmod{24}$ as $a(a\pm1)$ is divisible by $2$

Evidently, we don't need prime, any odd number not divisible by $3$ will satisfy this

lab bhattacharjee
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