Jtdylktuy

$x^4equiv1pmod{10}$ for $x$ and $10$ coprime, by Euler's theorem, since $varphi (10)=4$.

(The proof for $x$ not coprime to $10$ can be found in some of the other answers here.)

This is true because of the periodicity of the last digits of numbers raised to powers. if you check cor $x=2,3,4ldots n$ you will see that after every $4$th power the last digit is the same, $2^1 = 2$ and $2^5 = 32$ thus the last digit is the same. $x^n mod 10$ is essentially asking for the last digit. Thus, $$x^{n+4} mod 10 = x^n mod 10$$
Some numbers have a periodicity of $2$ when raised to powers, for example, $9$ but the unit digit is still the same as the first power as the first.

We want to show that $x^{n+4}-x^n=x^n(x^4-1)$ is a multiple of $10$.

We first notice that if $x^n$ is odd then $x^4-1$ will be even and vice versa. Consequently, the product will always be even.

If $x$ is divisible by $5$ then the product is divisible by $10$ because we have an even product which is divisible by $5$.

If $x$ is not divisible by $5$ then $x=5m+k$ for some $m$ and $k$ where $kin{1,2,3,4}$.

Now, $x^4-1=(5m)^4+4(5m)^3k+6(5m)^2k^2+4(5m)k^3+k^4-1$ and every term is divisible by $5$ with the possible exception of $k^4-1$.

But there are only four values of $k$ and if you raise each to the fourth power and subtract $1$ you get a multiple of $5$. Once again, this gives us an even number divisible by $5$ which is a multiple of $10$.

It is the special case $,p,q,k = 2,5,4,$ below.

Lemma $,pneq q,$ primes and $, color{#c00}{p!-!1,,q!-!1mid k} $ $Rightarrow, pqmidsmash[t]{overbrace{ x^n(x^k!-!1)}^{Large a}},$ for all $,x,$ and $,n>0$

Proof $ $ $,p,q,$ are coprime so $,pqmid aiff p,qmid a,,$ by Euclid / unique factorization.

When $, pmid x,$ then $,pmid xmid x^nmid a,$ by $,n>0,,$ hence $,pmid a,$ by transitivity of divisibility,

$ $ else: $, pnmid x,$ so $!bmod p!: xnotequiv 0,$ so $,x^k equiv (color{#0a0}{x^{p-1}})^{smash[t]{Largecolor{#c00}{frac{k}{p-1}}}}!!equivcolor{#0a0} 1,$ by $rmcolor{#0a0}{Fermat},,$ so $,pmid x^k!-1mid a$.

So in every case $,pmid a,,$ so $,qmid a,$ by $,p,q,$ symmetry (i.e. same proof works for $q). $

Remark $ $ Above is a special case of this generalization of Euler-Fermat - which often proves handy.

Theorem $ $ Suppose that $ min mathbb N $ has the prime factorization $:m = p_1^{e_{1}}cdots:p_k^{e_k} $ and suppose that for all $,i,,$ $ ege e_i $ and $ phi(p_i^{e_{i}})mid f. $ Then $ mmid a^e(a^f-1) $ for all $: ain mathbb Z.$

Proof $ $ Notice that if $ p_imid a $ then $:p_i^{e_{i}}mid a^e $ by $ e_i le e.: $ Else $:a:$ is coprime to $: p_i:$ so by Euler's phi theorem, $!bmod q = p_i^{e_{i}}:, a^{phi(q)}equiv 1 Rightarrow a^fequiv 1, $ by $: phi(q)mid f. $ Since all $ p_i^{e_{i}} | a^e (a^f - 1) $ so too does their lcm = product = $m$.

Examples $ $ You can find many illuminating examples in prior questions, e.g. below

$24mid a^3(a^2-1)$

$40mid a^3(a^4-1)$

$88mid a^5(a^{20}-1)$

$6pmid a,b^p - b,a^p$

If you can prove $x^4 equiv 1 pmod {10}$ that's enough as $x^{n+4}equiv x^ncdot x^4 equiv x^ncdot 1equiv x^npmod {10}$

If you google Euler's theorem that will be true for all numbers relatively prime to $10$. Intuitively (but hand wavy) if $x$ and $10$ are relatively prime then $x^n$ will be relatively prime. There are only $4$ possible last digits that are relatively prime to $10$ so $x^k$ will cycle through the same four digits. (More or less.)

If $x$ is even or divisible by $5$. Well, if it's both then $x equiv 0 mod 10$ so $x^k equiv 0 pmod {10}$ and $x^{n+4} equiv x^n equiv 0pmod {10}$. Likewise if it's odd but divisible by $5$ then $x^k equiv 5 pmod{10}$ and so $x^{n+4} equiv x^n equiv 5 pmod {10}$.

Now if $x$ is even but not divisible by $5$ then. Well, $x^n$ is even and there are only $4$ even digits it could end with and it cycles through them.

More to the point $2*j pmod 10 equiv 2*(j pmod 5)$ and there are only $4$ non-zero classes $pmod 5$ so $x^k$ just cycles through those.

Because:

$x^{n+4} = x^{n-1}x^5$
$x^n = x^{n-1}x$

($abmod 10) = (a(bmod 10) mod 10)$

... you only have to prove that $(x^5 mod 10)=(xmod 10)$.

... which means: $((x mod 10)^5mod 10) = (x mod 10)$.

Because there are only 10 possible values for $(x mod 10)$ these 10 values can be calculated directly:

• $0^5mod 10=0 mod 10=0$


• $1^5mod 10=1 mod 10=1$


• $2^5mod 10=32 mod 10=2$


• 
  $3^5mod 10=243 mod 10=3$
  
  ...


• $9^5mod 10=59049 mod 10=9$