Hi
I need help with the following question
I have attached my question.
I haven't got a clue on how to start the question.
thanks
You solve it by doing it! What is $\displaystyle 0^2 (mod 7)$? What is $\displaystyle 1^2 (mod 7)$? What is $\displaystyle 2^2 (mod 7)$? What is $\displaystyle 3^2 (mod 7)$? What is $\displaystyle 4^2 (mod 7)$? What is $\displaystyle 5^2 (mod 7)$? What is What is $\displaystyle 6^2 (mod 7)$?
Now reduce $\displaystyle 7x^5+ 3y^4= 2$ "(mod 7)". Since 7= 0 (mod 7) that reduces to $\displaystyle 3y^4= 3(y^2)(y^2)= 2$ (mod 7). Is there any integer y for which that is true? Replace $\displaystyle y^2$ with 0, 1, 2, and 4 to see.
Obviously:
$\displaystyle 7x^5+3y^4 \equiv 3y^4\ (\text{mod}\ 7)$
Now we know that $\displaystyle y^4\equiv 0, 1,2, \text{ or } 4\ (\text{mod} \ 7)$
So what are the possible values of $\displaystyle 3y^4\ (\text{mod}\ 7)$ and are any of these congruent to $\displaystyle 2 \ (\text{mod}\ 7)$
CB
It's the right hand side of your d**n problem:
$\displaystyle 7x^5+3y^4=2$
There is a bit of a hint also in the question that $\displaystyle \text{mod 7}$ is involved.
To prove that an equation has no integer solutions it is sufficient to show that it has no solution modulo some integer.
(As a hint in future read your own problem when you post it)
CB