# Number theory question involving primes [duplicate]

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.

• There seems to be a solution on AoPS, and some weaker proofs on this site Prime numbers and divisibility – Sil May 31 at 17:20
• Maybe the fact that a^4 - b^4 = (a^2 + b^2)(a+b)(a-b) could help. – Giacomo Maletto May 31 at 17:21
• Please use MathJax to format your posts – saulspatz May 31 at 17:22
• $a^4\equiv1\bmod16$ (Carmichael), $a^4\equiv1\bmod5$ (Fermat), and $a^2\equiv1\bmod3$ (Fermat) if $\gcd(a,240)=1$ – J. W. Tanner May 31 at 18:09
• This may be an even better duplicate target. – Jyrki Lahtonen Jun 1 at 5:02

$$240 = 2^4 \cdot 3 \cdot 5$$. 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 $$17^4-11^4$$?
• "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 $17^4-11^4$? – Robert Israel Jun 1 at 3:44
• $240$ 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! – Aayam Mathur Jun 1 at 3:51