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Theorem (Legendre): Let a,b,c coprime positive integers, then $ax^2+by^2=cz^2$ has a nontrivial solution in rationals x,y,z iff $(−bc/a)=(−ac/b)=(ab/c)=1$.

I read this somewhere. Is it really the way the theorem is stated? Doesn't it imply that there are are non-trivial solutions if $\gcd(a,b,c)=1$?

2 Answers2

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There is a problem with the signs, it should be $$\color{red}{ (bc|a) = (ac|b) = (-ab|c) = 1 } $$ the way you wrote the coefficients.

This has been posted often enough before. It works as advertised if all the numbers are primes. However, for composite numbers there is a big difference between saying $(G|H) = 1$ and saying that $G$ really is a quadratic residue $\pmod H,$ which requires that $(G|p) = 1$ for every prime $p$ that divides $H.$

So, here is an example where the Jacobi symbols do not tell enough of the story: $$ 5 x^2 + 41 y^2 = 21 z^2. $$ We are alright on two out of three, as $5,41$ are primes, and we do get, with $21 \cdot 41 = 861$ and $5 \cdot 21 = 105,$ $$ (861|5) = 1 \; \; \mbox{AND} \; \; (105|41) = 1. $$ It is also true that $$ (-205|21) = 1. $$ However, the reason this is true is that $$ (-205|3) = -1 \; \; \mbox{AND} \; \; (-205|7) = -1. $$ Well, $$ -205 \equiv 5 \pmod {21}. $$ The actual quadratic residues $\bmod {21}$ are $$ 0,1,4,7,9,15,16,18 $$ and simply do not include $5.$

Now, if there were to be any solution to $5 x^2 + 41 y^2 = 21 z^2$ in nonzero integers, there would be such a solution with $\gcd(x,y,z) = 1.$ However, you can check that it is necessary for both $x,y$ to be divisible by $3$ in order to get $5 x^2 + 41 y^2$ divisible by $3.$ Once this happens, $5 x^2 + 41 y^2$ is actually divisible by $9.$ Therefore $z$ must be divisible by $3,$ and $3 | \gcd(x,y,z).$ This contradicts the assumption that there is a solution in nonzero integers. The word for this is anisotropic. The quadratic form $5 x^2 + 41 y^2 - 21 z^2$ is anisotropic in the $3$-adic numbers $\mathbb Q_3.$ It is also anisotropic in the $7$-adic numbers $\mathbb Q_7.$

Here is a more optimistic one: $3x^2 + 5 y^2 = 17 z^2.$ $$\color{red}{ (85|3) = (51|5) = (-15|17) = 1 }, $$ and $x=2,y=1,z=1$ works.

Here is another one with all primes $3 \bmod 4,$ so there can be no doubt about any $\pm$ signs: $3x^2 + 7 y^2 = 19 z^2.$ Once again, $12+7=19.$

Note on finding one integer solution: on page 82 of CASSELS, if there are any solutions, there is at least one $(U,V,W)$ with $$ a U^2 + b V^2 + c W^2 < 4abc. $$ Finite search!

Apparently the version with the $\pm$ signs as above is pretty widespread, see Proof of Legendre's theorem on the ternary quadratic form Widespread but wrong.

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Will Jagy
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  • http://www.encyclopediaofmath.org/index.php/Legendre_theorem – Will Jagy Oct 24 '14 at 00:28
  • Thanks Will Jagy. I have another more or less similar question, please see if u can give me any further input. –  Oct 24 '14 at 00:56
  • @NumThcurious, if you mean your earlier question on finding all solutions, there is a standard procedure, but only after finding one integer solution. Then all rational solutions can be found by stereographic projection onto the ellipse $a u^2 + b v^2 = 1.$ – Will Jagy Oct 24 '14 at 01:00
  • Funny, I actually found the stereo graphic projection on the ellipse but I can not find another integer solution other than $(x,y)=( \sqrt{\frac{cz^2-by^2}{a}}, y)$. –  Oct 24 '14 at 01:07
  • @num, what are you trying to do, deal with the whole thing while keeping $a,b,c$ as variables? On page 82 of Cassels, it is proved that, if there are any integer solutions, there is one with very small entries; this can be found by a finite computer search with explicit bounds. – Will Jagy Oct 24 '14 at 01:25
  • Not really. I am looking for a form like the Pythagorean triple. –  Oct 24 '14 at 01:27
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eqn.: $a(x)^2+b(y)^2=c(z)^2$

Above has solution:

$a=2z^2-y^2$

$b=x^2+z^2$

$c=2x^2+y^2$

For $(x,y,z)=(3,2,5)$

After removing common factor we get:

$(a,b,c)=(23,17,11)$

$23(3)^2+17(2)^2=11(5)^2$

Sam
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