# Thread: Prove x^6+7y^3=2 has no solution

1. ## Prove x^6+7y^3=2 has no solution

Hi, I have the following problem in a course titled "Introduction to Abstract Algebra"... I am not quite sure if I got the question right though because it looks so easy.

"Prove that the equation

$x^6+7y^3=2$

has no solution with x and y being in Z (set of integers)"

I think that is it because $y=(2-x^6)^{1/3}/(7)^{1/3}$...but I don't know how to explain it :-/

2. Originally Posted by mbmstudent
Hi, I have the following problem in a course titled "Introduction to Abstract Algebra"... I am not quite sure if I got the question right though because it looks so easy.

"Prove that the equation

$x^6+7y^3=2$

has no solution with x and y being in Z (set of integers)"
If you know a bit of modular arithmetic then the quickest way to see this is to notice that $x^6\equiv1\pmod7$, so you can't have $x^6\equiv2\pmod7$.

3. Originally Posted by Opalg
If you know a bit of modular arithmetic then the quickest way to see this is to notice that $x^6\equiv1\pmod7$, so you can't have $x^6\equiv2\pmod7$.
Ah... simple yet elegant, I like it.

4. Originally Posted by Opalg
If you know a bit of modular arithmetic then the quickest way to see this is to notice that $x^6\equiv1\pmod7$, so you can't have $x^6\equiv2\pmod7$.
I agree that this is a very nice little piece of work. However, if I may, how did you know that $x^6\equiv1\pmod7$?

-Dan

5. Ah! Fermat's Little Theorem. I got it now.

-Dan

6. Originally Posted by Opalg
If you know a bit of modular arithmetic then the quickest way to see this is to notice that $x^6\equiv1\pmod7$, so you can't have $x^6\equiv2\pmod7$.
We must also mention the case: $x^6\equiv0\pmod7$ (because it's a condition in Fermat's Little Theorem )

Anyway , nice solution .