Prove that $x+x^{3}+x^{5}+x^{7}+x^{9}+x^{11}\equiv 1\mod{11}$ has no solution

Question

Prove that the following congruence has no solution on $\mathbb{Z}$

$$x+x^{3}+x^{5}+x^{7}+x^{9}+x^{11}\equiv 1\mod{11}$$

It is giving me hard times, some hints about it? Thank you

Not an elegant approach, but seems like you only have a short list of numbers to try. — David P, Feb 03 '19 at 20:27
I did it, but i believe it could exist a more teoretical approach :) — F.inc, Feb 03 '19 at 20:31

lhf · Answer 1 · 2019-02-03T20:28:40.840

2

Hint:

Write $ x+x^{3}+x^{5}+x^{7}+x^{9}+x^{11} = x(1+x^{2}+x^{4}+x^{6}+x^{8}+x^{10}) $.

Set $y=x^2$ and find the second sum using Fermat's theorem.

edited Feb 03 '19 at 20:28

answered Feb 03 '19 at 20:27

lhf

216,483

1

Fermat's little theorem – Peter Foreman Feb 03 '19 at 20:28
Eureka! Thank you :) – F.inc Feb 03 '19 at 20:32

score 1 · Accepted Answer · answered Feb 10 '19 at 22:49

It's a well known formula that $x^n-1=(x-1)(x^{n-1}+...+1)$. Therefore, we have that the LHS is $$x(x^{12}-1)=(x-1)x(x+1)$$ Where we have used Fermat's Little Theorem to conclude $x^{11}=x$. For the product of three consecutive integers to be $1$, we must have none of them to be a multiple of $11$, which eliminates some cases. The remaining are easy to check.

Prove that $x+x^{3}+x^{5}+x^{7}+x^{9}+x^{11}\equiv 1\mod{11}$ has no solution

2 Answers2