Find all x's that the number on the pic,

is a power of a prime

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# How would you solve this?

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Find all x's that the number on the pic,

is a power of a prime

that's the taylor expansion of 1/(1-x) so it's rather trivial

No it's not. Nowhere does it mention that [math]|x| < 1[/math] and the series is finite.

even if the series is finite, it's value it's incredibly close to 1/(1-x) because it goes to 7th order.

for example, if you want that to equal 7, you need to input x=6/7. if the series is not infinite, then x is approximately equal to 6/7.

try it, calculate how much you get if you input x=6/7

and if x>1, then x has to be too large for the 7th order to not work, I wouldn't try anything under x=100000

1/(1-7) = -1/6, 1+7+7^2+7^3+7^4+7^5+7^6+7^7 is nowhere near that lol.

>I wouldn't try anything under x=100000

I wouldn't try anything under x=100000

>I wouldn't try anything under x=100000

I wouldn't try anything under x=100000

>I wouldn't try anything under x=100000

I wouldn't try anything under x=100000

>I wouldn't try anything under x=100000

I wouldn't try anything under x=100000

you're retarded

still not a prime number, what's your point?

what part of

>and if x>1, then x has to be too large for the 7th order to not work, I wouldn't try anything under x=100000

do you not fucking understand?

>and if x>1, then x has to be too large for the 7th order to not work, I wouldn't try anything under x=100000

and if x>1, then x has to be too large for the 7th order to not work, I wouldn't try anything under x=100000

>and if x>1, then x has to be too large for the 7th order to not work, I wouldn't try anything under x=100000

and if x>1, then x has to be too large for the 7th order to not work, I wouldn't try anything under x=100000

>and if x>1, then x has to be too large for the 7th order to not work, I wouldn't try anything under x=100000

i've already given the answer, x=1 is the only value that works

8 is not a prime chud

is a prime power tho

It's the sum of a geometric series.

It is (1-x^8)/(1-x) with no approximation

factors as (x+1)(x^2+1)(x^4+1), so need x+1 = p^l, so x^2+2x+1 = p^(2l), where x^2+1 = p^m, so p^m = p^(2l) - 2p^(l) + 2, if l = 0 we get x = 0 which doesn't work, so l!=0, so p divids lhs and rhs, so p divides 2, get p = 2, but if m > 1 and l > 1 clearly 4 stops equality. m>=l, so take l = 1, we get x = 1 which doesn't work so done none work.

nvm x = 1 gets you 2^3, and i guess x = 0 = 1 is technically a prime power? rest is still valid

Incorrect.

Also incorrect.

It can be factored as (x^8 - 1)/(x - 1) so simply put: consider it as a geometric series, which leads to (x^8 - 1)/(x - 1) = (x^4 + 1)(x^2 + 1)(x + 1).

I forgot to mention that the x is greater than 0 and it's a whole number

but why x = 1 is the only answer?

Because this

The only value of x where [math](x^4 + 1)(x^2 + 1)(x + 1)[/math] is a power of a single prime is when x=1.

but where is it stated that it has to be a single prime, also how solving this equation can be a solution to this, please elaborate

> is a power of a prime

It's stated in the OP

> how solving this equation can be a solution to this

expand the expression and you get the OP equation, it's the same function.

but what's the logical reasoning behind this, how come two forms of the same function give us an answer

What? Are you retarded?

if x + 1 and x^2 + 1 are both powers of a prime p, then p divides 2x, so p divides 2 (p can't divide x) and p = 2. Then x is odd, say x = 2n + 1, so x^2 + 1 = 4n^2 + 4n + 2 = 2(2n^2 + 2n + 1), which is not a power of 2 unless n = 0 or -1, and if n = -1, x + 1 = 0. Hence, n = 0 and x = 1 can be verified by manually plugging it in.

actually, x = 0 is a solution.

1 is not a prime

illiterate retards

I believe that the retard is you anon.

Non sequitur.

1 is not a prime, so x=0 is not a solution.

it is a prime power tho

partial sum i think would work

x=3