1. Number Theory Help

I need help solving a problem for my discrete class.

The problem is : Prove no integer solutions of x and y satisfy the equation $\displaystyle 3x^2 + 5y^2 = 19$. Hint: Consider this equation in modulo 5.

My attempt:
Assume $\displaystyle 3x^2 + 5y^2 = 19$
Thus, in mod 5,

$\displaystyle 3x^2 + 5y^2 \equiv 19$

$\displaystyle 3x^2 + 5y^2 \equiv 4$
Note that 5y^2 in mod 5 will always be 0, since it will be a multiple of 5, thus:

$\displaystyle 3x^2 \equiv 4$

Now I broke it case by case to see what x^2 will have to be to satisfy the above requirement.
I found that

$\displaystyle x^2 \equiv 3$ in order for
$\displaystyle 3x^2 \equiv 4$

$\displaystyle x^2 \equiv 3$ implies $\displaystyle (\exists k \in Z )(5k+3 = x^2)$

The following steps may be useless and/or incorrectly derived, but i dont know what else to do:

if 5k+3 = x^2 , then $\displaystyle 5k = x^2 - 3$

What now?

2. Re: Number Theory Help

If $\displaystyle x \equiv 0 \pmod{5}$ then $\displaystyle x^2 \equiv 0 \pmod{5}$.
If $\displaystyle x \equiv 1 \pmod{5}$ then $\displaystyle x^2 \equiv 1 \pmod{5}$.
If $\displaystyle x \equiv 2 \pmod{5}$ then $\displaystyle x^2 \equiv 4 \pmod{5}$.
If $\displaystyle x \equiv 3 \pmod{5}$ then $\displaystyle x^2 \equiv 4 \pmod{5}$.
If $\displaystyle x \equiv 4 \pmod{5}$ then $\displaystyle x^2 \equiv 1 \pmod{5}$.

3. Re: Number Theory Help

there's probably a more formal method but you can notice that

$x^2 \pmod{5} = (x \pmod{5})(x \pmod{5}) \pmod{5}$

so we only have to look at $x \in \{0,1,2,3,4\}$

$x^2 \pmod{5} \in \{0,1,4\} \Rightarrow 3x^2 \pmod{5} \in \{0,2,3\}$

$4 \not \in \{0,2,3\}$

so the equation has no solutions mod 5

if there was a general solution it would be a solution mod any integer