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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.

Jyrki Lahtonen
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Aayam Mathur
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1 Answers1

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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 $17^4-11^4$?

Robert Israel
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  • Hmm i see.But how do i prove the gcd thing ( 240 is the gcd of all such numbers) in general? – Aayam Mathur Jun 01 '20 at 03:42
  • "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 01 '20 at 03:44
  • $240$ 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 01 '20 at 03:51
  • @robert israel How to solve this problem without using modular arithmetic. Because the book I'm referring has not introduced modular arithmetic yet. – mathophile May 27 '21 at 09:13