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Prove that, if a, b are prime numbers $a > b$, each containing at least two digits, then $a^4 - b^4$ is divisible by $240$. Also prove that, $240$ is the gcd of all the numbers which arise in this way.

Looking at the prime factorisation $240=(2^4)*3*5$, i know i need to prove that the given difference is divisible by each of these.

How do i proceed from here? i have no idea. Thanks.

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    $\begingroup$ There seems to be a solution on AoPS, and some weaker proofs on this site Prime numbers and divisibility $\endgroup$ – Sil May 31 at 17:20
  • $\begingroup$ Maybe the fact that a^4 - b^4 = (a^2 + b^2)(a+b)(a-b) could help. $\endgroup$ – Giacomo Maletto May 31 at 17:21
  • $\begingroup$ Please use MathJax to format your posts $\endgroup$ – saulspatz May 31 at 17:22
  • $\begingroup$ $a^4\equiv1\bmod16$ (Carmichael), $a^4\equiv1\bmod5$ (Fermat), and $a^2\equiv1\bmod3$ (Fermat) if $\gcd(a,240)=1$ $\endgroup$ – J. W. Tanner May 31 at 18:09
  • $\begingroup$ This may be an even better duplicate target. $\endgroup$ – Jyrki Lahtonen Jun 1 at 5:02
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$240 = 2^4 \cdot 3 \cdot 5$. Any prime $> 5$ is coprime to $2, 3, 5$. The fourth powers of odd numbers mod $2^4$ are all $1$, the fourth powers of $1$ and $2$ mod $3$ are $1$, and the fourth powers of $1,2,3,4$ mod $5$ are all $1$. So the fourth power of any number coprime to $240$ mod $240$ is $1$.

The first three two-digit primes are $11, 13, 17$.
What is the gcd of $13^4-11^4$ and $17^4-11^4$?

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  • $\begingroup$ Hmm i see.But how do i prove the gcd thing ( 240 is the gcd of all such numbers) in general? $\endgroup$ – Aayam Mathur Jun 1 at 3:42
  • $\begingroup$ "The gcd of all the numbers" is $240$ if all the numbers are divisible by $240$ and there are two whose gcd is $240$. Did you compute the gcd of $13^4 - 11^4$ and $17^4-11^4$? $\endgroup$ – Robert Israel Jun 1 at 3:44
  • $\begingroup$ $240$ is the $gcd$. Oh waitttt. Now for all other numbers formed this way, even if someof them had a gcd greater than $240$, the aggregate gcd would still be $240$. Understood. Thanks a lot! $\endgroup$ – Aayam Mathur Jun 1 at 3:51

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