elementary-number-theoryAug 21, 2026

How to prove that $x^{6n\pm1}+\frac{1}{x^{6n\pm1}}$ has the same units digit as $x+\frac{1}{x}$ ?

idrissbado

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I came across the following problem: If $x+\frac{1}{x}$ is an integer, prove that

x^{6n+1}+\frac{1}{x^{6n+1}}

and

x^{6n-1}+\frac{1} {x^{6n-1}}

have the same units digit as

x+\frac{1}{x},

for every positive integer $n$ . Equivalently, I need to prove that

x^{6n\pm1}+\frac{1}{x^{6n\pm1}} \equiv x+\frac{1}{x} \pmod{10}.

The only given condition is

x+\frac{1}{x}\in\mathbb{Z}.

Importantly, $x$ itself is not necessarily an integer. I tried defining

S_k=x^k+\frac{1}{x^k}.

Then, if

a=x+\frac{1}{x}\in\mathbb{Z},

we have the recurrence

S_{k+1}=aS_k-S_{k-1}.

This shows that every $S_k$ is an integer, but I am unable to prove the stronger result

S_{6n\pm1}\equiv S_1\pmod{10}.

What is the simplest way to prove this result?

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1 answer

idrissbado

Research member

Aug 21, 2026

Define

S_k(x)= x^k+\frac 1{x^k}

And set $S_1(x)= a$ . We remark that $S_2(x)=a^2-2$ .

We have

S_k(x)= \left(x^{k-1}+\frac 1{x^{k-1}}\right)\left(x+\frac 1x\right)-\left(x^{k-2}+\frac 1{x^{k-2}}\right)

whence the recursion

S_k(x)= aS_{k-1}-S_{k-2}

Note: we really only care about this $\pmod {10}$ .

It's now a simple matter to check each value of $a\pmod {10}$ . For instance, if $a\equiv 0 \pmod {10}$ , the $S_k(x)\pmod {10}$ are periodic with period $4$ and the sequence they form is

\overline {\{0,8,0,2\}}

(Note that all odd indexed numbers in the sequence end in $0$ which is stronger than we need).

Similarly, if $a\equiv 1 \pmod {10}$ we get the periodic block

\overline {\{1,9,8,9,1,2\}}

from which the relevant claim follows at once.

And similarly for the rest.

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