What is the maximum value of $n$ if $n + 5$ divides $n^5 + 5$, $n$ being a natural number? I tried to solve by binomial theorem but it failed.
$\max(\{n: n+5 | n^5 + 5\})$
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elementary-number-theory
divisibility
3 Answers
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If $n+5$ divides $n^5+5$, then it must also divide the remainder after dividing $n^5+5$ polynomially by $n+5$. This remainder will be a constant, meaning that the largest possible value of $n+5$ will be (the absolute value of) that constant.
By way of example, suppose we are expected to find the largest $n$ such that $n+3$ divides $n^3+3$. If we divide $n^3+3$ by $n+3$, we get
$$ \frac{n^3+3}{n+3} = n^2-3n+9-\frac{24}{n+3} $$
So $n+3$ must divide $24$, and the largest value of $n$ for which that can happen is $24-3 = 21$.
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Let $N=n+5$. Then
$$n^5+5=(N-5)^5+5=(N^5-\cdots-5^5)+5=N(something)-3120$$
so $N\mid n^5+5$ if and only if $N\mid3120$. The largest $N$ dividing $3120$ is $3120$ itself, so the largest $n$ is $3115$.
Remark: You can, if you like, write out all the terms in the binomial expansion of $(N-5)^5$, but it's really not necessary. What's important is that all but the final term, $-5^5$ are multiples of $N$.
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0+1 That's a good observation, that you don't need to expand $(n-5)^5$ (or, in the case of my answer, to divide out all the way). – 2017-02-07
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You have that $n+5 \mid n^5+5^5$ since $5$ is odd($a+b\mid a^{2k+1}+b^{2k+1}$). So you must have $n+5\mid n^5+5^5-n^5-5$ So $n+5\mid 5^5-5=3120$. It should be easy from here to conclude the max value of $n$.