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I am unable to get a start for this question

$2x^2+3y^2=z^2$ does not have a non-zero integral solution.

Can anyone please give me some hint so that I can try this , as I am unable to start.

Thanks

Sachchidanand Prasad
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    arguments using the prime 2 or the prime 3 will both work. What is your background relevant to the question? In particular, Legendre's theorem on indefiite (diagonal) ternary forms? – Will Jagy Dec 09 '16 at 22:50
  • I have read one course in number theory, but the thing is that I have tried this by observing that 2 and 3 are prime but unable to get anything. So could you give me a bit hint so that I can be procced. Thanks – Sachchidanand Prasad Dec 09 '16 at 22:52

3 Answers3

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Dividing by common factors, we can assume that $x,y,z$ have no common factors. Since $$ \begin{align} 2x^2&\equiv0,2&\pmod3\\ 3y^2&\equiv0&\pmod3\\ z^2&\equiv0,1&\pmod3 \end{align} $$ The only way we can have $2x^2+3y^2=z^2$ is if $x\equiv0\pmod3$ and $z\equiv0\pmod3$. However, this means that $9\mid3y^2$ and so $3\mid y$. However, this implies that $3$ divides each of $x,y,z$, which is a contradiction.

robjohn
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The hint would be to go modulo $3$. Hopefully, that will suffice! If not, here is a solution:

Assume the triple $(x, y, z) \neq (0, 0, 0)$ is reduced, i.e., has a gcd of $1$. Modulo $3$, though, it will follow that $3 ~|~ x, z$, and so $3 ~|~ y$, contradiction.

Triskele
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if $$ 2 x^2 + 3 y^2 - z^2 \equiv 0 \pmod 9, $$ then $x,y,z$ must all be divisible by 3.

However, if there is a solution to

$$ 2 x^2 + 3 y^2 - z^2 =0, $$ there is a solution with $$ \gcd(x,y,z) = 1. $$

Contradiction of existence with integers.

There is a power of $2$ that can be used instead, if $$ 2 x^2 + 3 y^2 - z^2 \equiv 0 \pmod 8, $$ then $x,y,z$ must all be divisible by $2.$

Will Jagy
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  • I don't see where this argument fails for $4x^2 + 3y^2=z^2$ (soln $x = 1, y= 2, z = 4$) – Mitch Dec 09 '16 at 23:03
  • Is there any particular reason your first congruence is mod $9$ instead of mod $3$? – Barry Cipra Dec 09 '16 at 23:23
  • @Mitch if $4x^2+3y^2=z^2$ then there are solutions where $x,z$ are not divisible by $3$ whereas $y$ is. Such solutions do not exist if the coefficient on $x^2$ is replaced by a number not $\equiv 1$ mod $3$. – Oscar Lanzi Dec 10 '16 at 00:35
  • @BarryCipra yes, it is possible to have the ternary form $0 \pmod 3$ with $x=0, y = 1, z=0,$ that is not all variables divisible by $3.$ The concept here is that the form is anisotropic in both $\mathbb Q_2$ and $\mathbb Q_3;$ it is not possible to have the form congruent to zero mod a high power of either prime without all three variables being divisible by by a high power ( about halfway) of that prime. see http://math.stackexchange.com/questions/27471/proof-of-legendres-theorem-on-the-ternary-quadratic-form – Will Jagy Dec 10 '16 at 00:56
  • @WillJagy, ah, of course! Thanks! – Barry Cipra Dec 10 '16 at 01:01
  • @OscarLanzi so then which $p,q,r$ allow integral solutions for $p x^2 + q y^2 = r z^2$. (1,1,1) obviously but... – Mitch Dec 10 '16 at 19:58
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    @mitch see https://www.encyclopediaofmath.org/index.php/Legendre_theorem – Will Jagy Dec 10 '16 at 23:02
  • @WillJagy Nice. Thanks. – Mitch Dec 12 '16 at 14:17