Prove that if $(x+y)^5$ is congruent to $2$ (mod $5$) then $(x+y)$ is congruent to $2$ (mod $5$)?

Question

I need a proof of the above. I understand that it is just an application of Fermat's Little Theorem, but how do I prove this without the theorem?

Answer 1

Use that if $n\equiv 0,1,2,3,4\mod 5$ then $n^5\equiv 0,1,2,3,4\mod 5$

Answer 2

You can just run through all the possible cases. Since you are not doing anything with $x$ or $y$ independently of the other, you can just assign $z = x + y$. And then there are only five cases, so it shouldn't take long to go through all the possibilities.

• $0^5 \equiv 0 \pmod 5$
• $1^5 \equiv 1 \pmod 5$
• $2^5 = 32 \equiv 2 \pmod 5$
• $3^5 = 243 \equiv 3 \pmod 5$
• $4^5 = 1024 \equiv 4 \pmod 5$

Another option is to do these calculations in base 5. But since $10 = 2 \times 5$, that doesn't offer much help.

Answer 3

$w^5 \equiv w \pmod 5$. Always.

Just calculate them. There are only five cases:

Cases 1-5: $w \equiv 0,1,2,3,4\pmod 5$ so $w^5 \equiv 0,1, 32, 243, 1024 \equiv 0,1,2,3,4 \pmod 5$.

So $(x+y)^5 \equiv x+y\pmod 5$ and if $(x+y)^5\equiv 2\pmod 5$ then $ x+y \equiv 2 \pmod 5$ is a peculiar specific application of it.

Of course we don't have to calculate them all. As $5$ is prime Fermat's Little Theorem say explicitely that $w^5 \equiv w\pmod 5$ and so their is nothing to prove at all.

Answer 4

Hint $$x^5-x =x(x^2-1)(x^2+1) \equiv x(x^2-1)(x^2+1-5)=(x-2)(x-1)x(x+1)(x+2) \pmod{5}$$

Now use the fact that among 5 consecutive integers one is divisible by $5$.