# Help Proving Conjecture?

Gold Member
Hello;

I don't know how to prove this conjecture I've made;

For the polynomial equation

$$-\sum_{n=0}^{\infty}k^{n}=0$$

there exist no real solutions greater than 2, no matter how large the value of n.

How do I show that this is true?

If it's a little unclear, what I mean is, for example, if you had the equation

k^30 - k^29 - k^28 ... - k - 1 = 0, k doesn't exceed 2.

So even if you start from 1000, or 100000000000, or any number, you will never find a solution greater than 2.

Thanks.

Staff Emeritus
Homework Helper
Homework Helper
I am assuming your latex equation has something missing, because otherwise the minus sign at the start doesn't have any purpose. So working from your example equation:

$$k^n - k^{n-1} - k^{n-2} \cdots - k - 1 = 0$$

Multiply by k:

$$k^{n+1} - k^{n} - k^{n-1} \cdots - k^2 - k = 0$$

Subtract the first equation from the second:

$$k^{n+1} - 2k^n + 1 = 0$$

$$k^n(k-2) + 1 = 0$$

Which has no solution if k > 2

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