Thread: 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.

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 ,
(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 ?

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 has no integer non-negative solutions.

From we conclude, by first checking what a square can be modulo 5,
that and .

Put this in the equation, check that and replace with .

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).

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 has no integer non-negative solutions.

From we conclude, by first checking what a square can be modulo 5,
that and .

Put this in the equation, check that and replace with .

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