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 , x˛ 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).