0

I have been trying to share information concerning a conjecture that for all Prime numbers greater than 3, the difference between the their squares, modulus 24, is zero. I have tested this for the first 6 million primes without a contrary example. I am continuing the search but this is extremely time consuming.

I realise that this information is trivial and only worthy of the briefest of notes, and I cannot find where to pass on the information. Does anyone know where or who I can ask whether this is a known conjecture?

Maurice
  • 9
  • 2
    It is indeed known and has been asked here already a few times. You can search this site for such results. – Dietrich Burde Mar 10 '21 at 14:57
  • Let $p,q$ be primes, both greater than $3$ with $p>q$. Then $(p^2-q^2)=(p-q)(p+q)$. Clearly, this is divisible by $3$ since $3\nmid q$ it follows that among $p-q,p,p+q$ one must be divisible by $3$ and it couldn't have been $p$. Further, as $2\nmid q$ among $p-q,p,p+q$ one will be divisible by $2$ with another divisible by $4$. As it again cannot be $p$ we get that $(p-q)(p+q)$ is divisible by $8$. As $(p-q)(p+q)$ is divisible by both $3$ and $8$ it is therefore divisible by $24$. – JMoravitz Mar 10 '21 at 14:59
  • 3
    It seems like a routine elementary-number-theory exercise in divisibility, not really the type of thing that remains an open conjecture for long. – JMoravitz Mar 10 '21 at 15:00
  • This is in fact a trivial exercise. It is well known that every odd integer $n$ satisfies $n^2\equiv 1\mod 8$ (or it can easily be shown) and $n^2\equiv 1\mod 3$ is also trivial if $3\nmid n$. This shows that every integer $n$ with $\gcd(n,6)=1$ satisfies $n^2\equiv 1\mod 24$. This immediately implies the claim. – Peter Mar 11 '21 at 18:41
  • Even more elegant is to use the Carmichael-function : $\lambda(24)=2$ , hence $n^2\equiv 1\mod 24$ holds for every $n$ that is coprime to $24$ which is equivalent to be coprime to $6$. – Peter Mar 11 '21 at 18:47

0 Answers0