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Find all positive integers $n$ such that $n^4 − 1$ is divisible by 5.

I want help with this problem. I have tried using factorization to $(n-1)(n+1)(n^2+1)$. but do not know how to proceed further. I think my approach of factorization is not good.

user21820
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Deepak Lohana
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    Your factorization is correct. Now, notice that $5$ is prime. $5\mid a\times b\times c$ is true for integers $a,b,c$ if and only if $5\mid a$ or $5\mid b$ or $5\mid c$. Can you continue? – JMoravitz Oct 28 '19 at 19:26

3 Answers3

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Hint:

As $5$ is prime, you can use lil' Fermat:

For all numbers $n$ not divisible by a prime $p$, one has $\;n^{p-1}\equiv 1\mod p$.

Bernard
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  • @DietrichBurde:He was rather at the opposite side. of France. It's just a nickname I usually give, in reference to a famous comic strip from the thirties (lil' Abner) because one of my friends used to call it ‘petit Fermat’. – Bernard Oct 28 '19 at 20:07
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every such positive integer that is not a multiple of $5$ satisfies the constraint. Consider quadratic residues modulo $5.$ Every integer is of the form $5k,5k+1,5k+2,5k+3,5k+4,$ where $k$ is an integer. If you substitute all the forms other than $5k$ into the equation, then you get a multiple of $5$ each time. Note that you only need to consider the constant term as all nonconstant terms will be multiples of $5,$ which is easy to see using the binomial theorem.

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We have that for any $n \implies n \equiv 0,1,2,3 \,\text{or}\,4 \mod 5 $ and

  • $1^4-1\equiv 0 \mod 5$

  • $2^4-1\equiv 0 \mod 5$

  • $3^4-1\equiv 0 \mod 5$

  • $4^4-1\equiv 0 \mod 5$

user
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