Using Fermat's Little Theorem to Show Divisibility

Question

I was asked to prove, using Fermat's Little Theorem, that $11|5^{10n+8}-4$ for $n\ge0$. I proved it but I was wondering whether there's an easier way (still using Fermat's). Here is my proof:

\begin{alignat}{3} 11|5^{10n+8}-4&\iff5^{10n+8}-4&&\equiv0 &&&\mod11\\ \quad&\iff 25^{5n+4}-4&&\equiv0 &&&\mod 11\\ \quad&\iff \qquad3^{5n+4}&&\equiv 4 &&&\mod 11\\ \quad&\iff \qquad3^{5n+5}&&\equiv 12 &&&\mod 11\\ \quad&\iff \qquad3^{5(n+1)}&&\equiv 1 &&&\mod 11.\\ \end{alignat} For $n\ge1$, let S(n) be the statement

$$ S(n) :3^{5(n+1)}\equiv 1 \mod 11.$$ We will prove by induction on $n$ that $S(n)$ holds.

Base case ($n=1$). By Fermat's Little Theorem, $S(1)$ is true.

Inductive Step. Fix some $k\ge1$ and suppose $S(k)$ is true. To be shown is that the statement $$S(k+1):3^{5(k+2)}\equiv 1 \mod 11$$ follows. Beginning with the LHS of $S(k+1)$,

\begin{alignat}2 \quad&3^{5(k+2)}&&=3^{5(k+1)+5}\tag{1}\\ \quad&\ \implies &&=3^{5}3^{5(k+1)}\tag{2}\\ \quad& \overset{\text{IH}}{\implies} &&\equiv3^{5}(1)\mod 11\tag{3}\\ \quad&\ \implies &&\equiv1\mod 11\tag{4},\\ \end{alignat} arriving to the RHS of $S(k+1)$, concluding the inductive step. It is proved, then, by MI that $S(n)$ holds for all $n\ge1.$ Since $S(0)$ holds by $(4)$, then $S(n)$ is true for all $n\ge0$.

This is far too complicated. Note that $5^8\equiv4\mod{11}$, and that by Fermat's Little Theorem, $5^{10}\equiv1\mod{11}$. The rest follows. — Rushabh Mehta, Dec 07 '19 at 20:22

score 6 · Accepted Answer · answered Dec 07 '19 at 20:50

6

We have $$ 5^{10n+8} = 5^{10n} 5^8 = (5^{10})^n 5^8 \equiv 1 \cdot 5^8 \equiv 4 \bmod 11 $$

answered Dec 07 '19 at 20:50

lhf

221,500

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From a comment by Don Thousand. – lhf Dec 07 '19 at 20:50
In particular, repeated squaring gives $5^2=3,,5^4=-2,,5^8=4$. – J.G. Dec 07 '19 at 20:54

score 3 · Answer 2 · edited Oct 20 '24 at 06:30

3

multiply by $25$ getting $5^{10(n+1)}-100$
Take remainders using Fermat, getting $1-1\equiv 0\pmod{11}$

edited Oct 20 '24 at 06:30

Bill Dubuque

282,220

answered Dec 07 '19 at 22:46

+1 That's one way I'd show it too. Worth emphasis is that $\bmod 11!!:,\ 25 x\equiv 0 \iff x\equiv 0\ $ by $\ 25,$ is invertible (so cancellable). – Bill Dubuque Dec 10 '19 at 05:54
It's not mod for that first part simply $p|a\implies p|ab$ ... – Dec 10 '19 at 11:10
Of course we can rewrite congruences in divisibility language, but generally that's not a good thing to do since it often obfuscates innate arithmetical structure. – Bill Dubuque Dec 10 '19 at 15:38
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I didn't really use much ... – Dec 10 '19 at 15:56
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To finish requires your way you could justify the inference $\ 11\mid 25x ,\Rightarrow, 11\mid x,,$ which is the divisibility form of what I wrote above. We can't possibly know what you intended to "use much" since you did not finish the argument, i.e. you omitted that step. – Bill Dubuque Dec 10 '19 at 16:18
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I honestly don't care bill. I'm not here to teach advanced versions of crap all. – Dec 10 '19 at 16:23
It has little to do with "advanced". Proofs need to be rigorous and complete at all levels. The point of my initial comment was to remind readers that this argument depends on that inference. Failing to recognize that will quickly lead one astray when working mod composites, e.g. $\bmod 10!:\ 2x\equiv 0\not\Rightarrow, x\equiv 0,,$ e.g. $,x = 5\ \ $ – Bill Dubuque Dec 10 '19 at 16:29
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It would only be advanced in my country as of the time I graduated, we had barely learned sequences of difference. – Dec 10 '19 at 16:32

score 3 · Answer 3 · edited Dec 08 '19 at 06:42

Much easier way!

By FLT $5^{10} \equiv 1 \pmod{11}$ so $5^{10n+8}\equiv 5^8$ and $5^{10n +8} -4 \equiv 5^8 -4\pmod {11}$.

So you just have to show that one case the $5^8 \equiv 4 \pmod {11}$. Then every case will be $5^{10n + 8} - 4\equiv 0 \pmod{11}$

Admittedly that requirse calculations but there are 3 ways, each more clever than the other

1) $5^2 = 25\equiv 3 \pmod {11}$. $5^4\equiv 3^2 \equiv 9\equiv -2 \pmod {11}$. $5^8\equiv (-2)^2 \equiv 4 \pmod {11}$.

2) $5^8*5^2 \equiv 5^{10} \equiv 1\pmod {11}$

$5^8*5^2 \equiv 5^8*3 \equiv 1\pmod{11}$ so as $11$ is prime $3^{-1}$ exist as is.... $1 \equiv 12=3*4\pmod{11}$ so $5^8*3*4 \equiv 4\pmod {11}$ and $5^8\equiv 4\pmod {11}$.

3) I'll admit I didn't come up with this.

If $5^8 -4 \equiv A\pmod{11}$ then

$(5^8-4)*25 \equiv A*25\pmod{11}$

$5^{10} - 100 \equiv 3A$

$1 - 1 \equiv 3A$

$3A \equiv 0\pmod {11}$ and as $11$ is primes $A\equiv 0 \pmod{11}$.

persistence of 0 Mod, modulo multiples for the win. – Dec 08 '19 at 11:22 — , Dec 08 '19 at 11:22

score 2 · Answer 4 · edited Dec 07 '19 at 23:47

2

$$10n+8=10(n+1)-2$$ we know that $5^{10}\equiv 1 \mod 11$ by Fermat.

we just need to prove that $$5^{-2}-2^2=$$ $$(5^{-1}+2)(5^{-1}-2)\equiv 0 \mod 11$$

which is true since the inverse $5^{-1}$ is $9$ .

edited Dec 07 '19 at 23:47

J. W. Tanner

63,683
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answered Dec 07 '19 at 20:59

hamam_Abdallah

63,719

score 1 · Answer 5 · answered Dec 07 '19 at 23:53

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$10\equiv-1\bmod11,$ so $10^8 \equiv(-1)^8=1\bmod11,$

so $5^{10n+8}\equiv5^8\equiv5^82^{10}=10^82^2\equiv4\bmod 11,$

since $5^{10}$ and $2^{10}\equiv1\bmod11$ by Fermat's little theorem.

answered Dec 07 '19 at 23:53

J. W. Tanner

63,683
4
43
88

Using Fermat's Little Theorem to Show Divisibility

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