# Why must any solution of this congruence equation satisfy gcd(x,11)=1?

Solve the congruence $$x^5\equiv 3 \space mod \space 11$$

The solution says that any solution of this equation must satisfy gcd(x,11)=1 and I'm not sure why.

I understand the theorem : "the linear diophantine equation $$ax+by=c$$ has solutions iff $$gcd(a,b)|c$$" but in this case the congruence is not linear...

Is it perhaps to do with the fact that if $$gcd(x,11) >1$$ then the gcd would have to be 11 and that would mean that x is divisible by 11 and thus $$x \equiv 0\space mod \space 11$$ and thus we would have $$(0)^5 \equiv 3 \space mod \space 11$$ and in this case we would of course not have any solution?

• Completely correct. – Parcly Taxel Jan 13 at 9:57
• The afirmation is true – El borito Jan 13 at 10:07

As you said, if $$\gcd(x, 11)>1$$ then it is $$11$$ and if $$11|x$$ it is easy to see that $$x^5$$ is congruent with $$0$$ $$\mod {11}$$