Show that $a^{25} \pmod{88}$ is congruent with $ a^{5} \pmod {88}.$

Question

Show that $a^{25} \pmod{88}$ is congruent with $ a^{5} \pmod {88}.$

I have proved it in the case that $\gcd(88,a)=1$, but in the other case , I don't know it. Any ideas?

Answer 1

As $(ab)^{25}\equiv (a^{25}b^{25})\pmod{88}$, and you have the result for numbers that are relatively prime to $88$, you only need to look at the prime factors of $88=2^3\cdot 11$. $11^{25}$ is a number some calculators might find too big, but apart from that technicality it should be really easy.

Answer 2

Note that $88=11 \cdot 8$, so $\Bbb Z _{88} \simeq \Bbb Z _8 \times \Bbb Z _{11}$. Therefore, you must show that $a^{25} = a^5$ in both $\Bbb Z _8$ and $\Bbb Z _{11}$. We shall use Euler's theorem: in $\Bbb Z _n$, we have $a^{\varphi (n)} = 1$, where $\varphi$ is Euler's function. Note that $\varphi(8)=4$ and $\varphi (11)=10$.

In $\Bbb Z _{11}$, every $a \ne 0$ is coprime to $11$, so $a^{10} = a^{\varphi (11)} = 1$, so $a^{25} = a^{20+5} = (a^{10})^2 a^5 = a^5$.

In $\Bbb Z _8$, if $\gcd (a, 8) = 1$, then $a^4 = a^{\varphi (8)} = 1$, so $a^{25} = a^{24+1} = (a^4)^6 a = a$; on the other hand, $a^5 = a^{4+1} = a^4 a=a$, so $a^{25}=a^5 \mod 8$.

If $\gcd(a, 8) >1$ then $2|a$, say $a=2b$. Then $a^{25} = 2^{25} b^{25} =0$, because $8 | 2^{25}$. Similarly, $a^5 = 2^5 b^5 = 0$, so again $a^{25} = a^5$.

The last case to examine is $a=0$, but this is trivial.

Answer 3

Since $\phi(8)=4$, if $(2,a)=1$, then $$ a^5\equiv a^{25}\pmod8\tag{1} $$ However, if $2\mid a$, then $a^5\equiv0\pmod8$ and $a^{25}\equiv0\pmod8$, so $(1)$ still holds.

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Since $\phi(11)=10$, if $(11,a)=1$, then $$ a^5\equiv a^{25}\pmod{11}\tag{2} $$ However, if $11\mid a$, then $a^5\equiv0\pmod{11}$ and $a^{25}\equiv0\pmod{11}$, so $(2)$ still holds.

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Putting together $(1)$ and $(2)$ gives $$ a^5\equiv a^{25}\pmod{88}\tag{3} $$

Answer 4

As $88=8\cdot11$

Using Fermat's Little Theorem, $11|(a^{11}-a)=a(a^{10}-1)\forall a\in \Bbb Z$

Again if $2|a,8|a^n$ for integer $n\ge3$

Else using $2\nmid a\iff(a,2)=1\implies a$ is odd $=2b+1$(say)

$(2b+1)^2=8\dfrac{n(n+1)}2+1\equiv1\pmod8\implies a^2\equiv1$

So, $8|a^3(a^2-1)\forall a\in \Bbb Z$

$\implies88|a^{\text{max}(1,3)}(a^{\text{lcm}(2,10)}-1)=a^3(a^{10}-1)$

Now $$a^{25}-a^5=a^5(a^{20}-1)$$ is divisible by $a^3(a^{10}-1)$

Answer 5

Apply the Theorem below, which is a generalization of Fermat and Euler's theorem $\rm\color{blue}{(E)}.$

From below $\ n^{20+5}\equiv n^{5}\!\pmod{2^3 11}\,$ by $\,2\neq 11\,$ are prime, $\ \color{#0a0}{4,10\mid 20},\ $ and $\ 3,1<5$

Theorem $\ \ \, n^{\large \varphi+k}\equiv n^{\large k}\pmod{p^i q^j}\ \ $ if $\ p\ne q\,$ are prime, $ \ \color{#0a0}{\varphi(p^i),\varphi(q^j)\mid \varphi},\ $ $\, i,j \le k $

${\bf Proof}\,\ \ p\nmid n\,\Rightarrow\, {\rm mod\ }p^i\!:\ n^{ \varphi}\equiv 1\,\Rightarrow\, n^{\varphi+k}\equiv n^k,\ $ by $\,\ n^{\large \color{#0a0}\varphi} = (n^{\color{#0a0}{\large \varphi(p^{\Large i})}})^{\large \color{#0a0}\ell}\overset{\color{blue}{\rm (E)}}\equiv 1^{\large \ell}\equiv 1$

$\qquad\quad\ \ \color{#c00}{p\mid n}\,\Rightarrow\, {\rm mod\ }p^i\!:\ n^k\equiv 0\,\equiv\, n^{\varphi+k}\ $ by $\ n^k = n^{k-i} \color{#c00}n^i = n^{k-i} (\color{#c00}{mp})^i$ and $\,k\ge i$

So $\ p^i\!\mid n^{\varphi+k}\!-n^k.\,$ By symmetry $\,q^j$ divides it too, so their lcm $ = p^iq^j\,$ divides it too. $\ $ QED

See also Carmichael's Lambda function, a generalization of Euler's phi function.