find n such that x^(8n + 3)=x^(n+1) is divisible by 23 for any integer x

Question

anonymous · Answer

$$x ^{8n+3} = x^{n+1}$$ the equal sign should be 3 bars not 2

anonymous · Answer

congruent in what sense?  Is it modulus?

anonymous · Answer

frankly to be honest I haven't got a clue...that's all the question said
this is the start of the question I think. the = is congruent and comma start of a new line
$$x ^{8n +3} = x^{n+1}   (\mod 23), x^{8n+3-(n+1)}=1, x^{7n}=1$$

ganeshie8 · Answer

u may use $\equiv$ for congruence..

anonymous · Answer

so I imagine that x^7n must be congruent with 1 (mod 23)

anonymous · Answer

$$
x^{8n+3}= x^{7n+2}x^{n+1}
$$

anonymous · Answer

dunno if that gets you anywhere

ganeshie8 · Answer

$\mathbb{x ^{8n +3} \equiv   x^{n+1}   (\mod 23)}$

$\mathbb{x ^{7n+3} \equiv   x   (\mod 23)}$

By Fermat,
$\mathbb{7n+3 =  23}$

$\mathbb{\implies  n =   20/7}$

however thats not a integer. fail

ganeshie8 · Answer

from which chapter is this question from ?

anonymous · Answer

n doesn't have to be an integer x does I think

ganeshie8 · Answer

for divisibility+fermat all needs to be integers. not just x

anonymous · Answer

okay...this is actually my homework so I have no clue...I don't get my lecturer whatsover yet somehow i managed to get most of the homework done.

ganeshie8 · Answer

this is a number theory problem. may i knw from which chapter/mopule this belongs to ?
that info may help in approaching the problem ..

anonymous · Answer

could it be something in encoding???? and perhaps would euclids help???? but really i am not sure

ganeshie8 · Answer

Actually you're right $\large n = \frac{20}{7}$ works perfectly fine here.

it doesnt make the exponent of x a fraction cuz 7 cancels out ( 7n)..