# How would you solve this? Find all x's that the number on the pic,
is a power of a prime

1. 2 weeks ago
Anonymous

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

• 2 weeks ago
Anonymous

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

• 2 weeks ago
Anonymous

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

• 2 weeks ago
Anonymous

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

• 2 weeks ago
Anonymous

>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

• 2 weeks ago
Anonymous

you're retarded

• 2 weeks ago
Anonymous

>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

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

• 2 weeks ago
Anonymous

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

• 2 weeks ago
Anonymous

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

• 2 weeks ago
Anonymous

8 is not a prime chud

• 2 weeks ago
Anonymous

is a prime power tho

• 2 weeks ago
Anonymous

It's the sum of a geometric series.

• 2 weeks ago
Anonymous

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

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

2. 2 weeks ago
Anonymous

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.

• 2 weeks ago
Anonymous

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

• 2 weeks ago
Anonymous

Incorrect.

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

Also incorrect.

3. 2 weeks ago
Anonymous

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).

4. 2 weeks ago
Anonymous

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

5. 2 weeks ago
Anonymous

but why x = 1 is the only answer?

• 2 weeks ago
Anonymous

Because this

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).

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.

• 2 weeks ago
Anonymous

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

• 2 weeks ago
Anonymous

> 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.

• 2 weeks ago
Anonymous

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

• 2 weeks ago
Anonymous

What? Are you retarded?

• 2 weeks ago
Anonymous

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.

• 2 weeks ago
Anonymous

actually, x = 0 is a solution.

• 2 weeks ago
Anonymous

1 is not a prime

• 2 weeks ago
Anonymous

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

illiterate retards

• 2 weeks ago
Anonymous

I believe that the retard is you anon.

• 2 weeks ago
Anonymous

Non sequitur.

• 2 weeks ago
Anonymous

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

• 2 weeks ago
Anonymous

it is a prime power tho

• 2 weeks ago
Anonymous
6. 2 weeks ago
Anonymous

partial sum i think would work

7. 2 weeks ago
Anonymous

x=3