1. ## diophantine equation

I have to show that the Diophantine eqn.

x^2 + 10y^2 = 2z^2 has no solutions

first i let x,y,z be a solution

then i got x^2 is congruent to 0 (mod 2)

but i don't know how this helps or any other way to approach it.

2. Did you try Fermat's Infinite Descent ?
mmm , sorry after trying it , i don't think it's a good way to treat this exercice .
But we can also see that this equation is equivalent to this one ,$4x'^2+10y^2=2z^2\Leftrightarrow 2x'^2+5y^2=z^2$
(because , as you said , x² is congruent to 0 (mod 2) so x is also congruent to 0 (mod 2) , then , x=2x' ...)
I also think that using inequalities is a good way ..... (i'll give it a try later )

P.S: Do we have to solve it in |N or Z ?

3. The only solution is (0,0,0) (over both the integers and the naturals).

Sketch of proof:

Since we're dealing with squares here, we can assume the integers are non-negative.

From what Tarask did, we're left with showing that $2x^2+5y^2=z^2$ has no integer non-negative solutions.

From $z^2\equiv 2x^2 \pmod{5}$ we conclude, by first checking what a square can be modulo 5,
that $5\mid x$ and $5\mid z$.

Put this in the equation, check that $5\mid y$ and replace $y$ with $5y'$.

Now we have the same equation as before, and thus we can conclude by Fermat's infinite descent that the only

solution is (0,0,0).

4. Originally Posted by Unbeatable0
The only solution is (0,0,0) (over both the integers and the naturals).

Sketch of proof:

Since we're dealing with squares here, we can assume the integers are non-negative.

From what Tarask did, we're left with showing that $2x^2+5y^2=z^2$ has no integer non-negative solutions.

From $z^2\equiv 2x^2 \pmod{5}$ we conclude, by first checking what a square can be modulo 5,
that $5\mid x$ and $5\mid z$.

Put this in the equation, check that $5\mid y$ and replace $y$ with $5y'$.

Now we have the same equation as before, and thus we can conclude by Fermat's infinite descent that the only

solution is (0,0,0).
I knew that we have to use FID , nice proof