$$3x^2 + 2y^4 = z^4$$

How do I solve this?? I would like to use so-called "elementary number theory", not abstract algebra (e.g. $\mathbb{Z} ( \sqrt d)$) or elliptic curves.

Note: I'm not asking what the solutions are, but rather how to find them.

My instincts are:

  • search the internet (I compared this equation with the ~280 here on MSE, and tried a variety of similar searches on uniquation.com ...)
  • search the 3 number theory books that I have
  • try to find solutions "by inspection" (possibly after reducing the order of the variables)
  • do some magic with modular arithmetic
  • use Alpern's solver - which seemed to indicate that there are no solutions (though I might have made an illegal substitution, so to speak)

I was able to identify $A = 6, B = 3, C = 6$ as solutions of $ \ 3A + 2B \ ^2 = C \ ^2$, but those aren't squares!

What is the number-theoretic approach to such problems? Is there a general method?

  • $\begingroup$ For your simplified problem, you should have kept the $x^2,$ with letter $U,V,W$ making $3 U^2 + 2 V^2 = W^2.$ This also has only the trivial solution $0,0,0.$ It is just quadratic residues, as in my answer below. $\endgroup$
    – Will Jagy
    Apr 25, 2012 at 18:50

4 Answers 4


Supposing we did have a solution lets consider the equation modulo $3$, since a square (hence a fourth power) must be congruent to $0$ or $1$ so the LHS is congruent to $0$ or $2$ and the RHS is $0$ or $1$ we see $3$ must divide both $y$ and $z$ thus $3^3$ must divide $x^2$ so $3^2$ divides $x$ hence $3^4$ divides the entire equation and dividing through leaves the same equation as we started with so we fall into an infinite decent which is absurd, hence we can have no solution in integers.

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    $\begingroup$ Very nice. One can also argue about the powers of 2 dividing the variables, and come to a similar contradiction. $\endgroup$ Apr 22, 2012 at 4:40
  • $\begingroup$ Dayo, this is definitely a solution, but I wonder: how did you decide to go mod 3?? What if the coefficient had been $7, 11, 300$, etc? What went through your head as you were trying to solve this? $\endgroup$ Apr 22, 2012 at 13:22
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    $\begingroup$ I can't speak for Dayo, but first of all there aren't all that many squares and 4th powers mod 3 and mod 9 (ditto mod small powers of 2) so if you're going to try congruences it makes sense to start with these. And having 2 and 3 as coefficients seals the deal. $\endgroup$ Apr 25, 2012 at 3:50
  • $\begingroup$ Thanks @Gerry. Not sure how I missed this notification until now. I might edit the question soon to further the discussion of methodology. $\endgroup$ Apr 25, 2012 at 19:53

Maybe what you need is Legendre's Theorem. Certainly it covers this situation. It tells you exactly what needs to be checked. It is presented in Ireland and Rosen, A Classical Introduction to Modern Number Theory, chapter 17, section 3. A very similar treatment is in PETE, pages 5-8.

Anyway, if $a,b,c$ are integers, not all positive and not all negative, but all nonzero, they are all squarefree, $\gcd(b,c) = \gcd(c,a) = \gcd(a,b) = 1,$, then $$ a x^2 + b y^2 + c z^2 = 0 $$ has a nontrivial solution in integers if and only if all three of these are true:

(i) $-bc$ is a square $\pmod a,$

(ii) $-ca$ is a square $\pmod b,$

(iii) $-ab$ is a square $\pmod c.$

In these we include $0$ as a square.

This result is also done in any book on quadratic forms. I like Rational Quadratic Forms by J. W. S. Cassels.

Anyhoo, you have $$ 3 U^2 + 2 V^2 - W^2 = 0,$$ so $a=3, \; b=2, \; c = -1.$ Condition (iii) asks if $-6$ is a square $\pmod {-1},$ the answer is yes as it is a multiple. Condition (ii), is $3$ a square $\pmod 2,$ again the answer is yes. However, condition (i), $2$ is not a square $\pmod 3.$ So there you are.

Leonard Eugene Dickson devoted a whole chapter to this in Introduction to the Theory of Numbers (1929). He still gave a section on it in Modern Elementary Theory of Numbers (1939).

NOTE: so this problem can be done without the fourth powers. The traditional one where the fourth powers really matter is $X^4 + Y^4 = Z^2,$ which is part of the proof of Fermat's Last Theorem, and is not just an application of Legendre's Theorem (although not difficult).


EDIT: it would appear that what you want to know a procedure. So, put all the degree four terms together on one side of the equals sign and re-write that as a quadratic form in substitute variables, as below.

The reason you use 3 is this: write $$ 3 x^2 = u^2 - 2 v^2, $$ where we will eventually put back $u = z^2, v = y^2.$

Now see my answer at Solving a Diophantine Equation

The discriminant of the binary quadratic form $u^2 - 2 v^2$ is $8.$ As $3$ is odd we can calculate $(8|3) = -1.$ From $$ u^2 - 2 v^2 \equiv 0 \pmod 3 $$ we find that $$ u,v \equiv 0 \pmod 3. $$

That is the quadratic forms part. When you finally put back $u = z^2, v = y^2$ you find that $$ 3 x^2 \equiv 0 \pmod {81}, \; \; x^2 \equiv 0 \pmod {27}, \; \; x \equiv 0 \pmod 9.$$ And so on.

EXTRA CREDIT: Try $$ 5 x^2 = 2 y^4 + y^2 z^2 + 3 z^4. $$ First, what is the discriminant of $2 u^2 + u v + 3 v^2?$

  • $\begingroup$ "discriminant of the binary quadratic form..." is just a highbrow way of saying 2 isn't a square mod 3. $\endgroup$ Apr 25, 2012 at 3:59
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    $\begingroup$ @Gerry, my brows are in the ordinary position. Also there are two of them. However, these days they do get these points which i think of as Wrath of Khan. I'm looking for a picture of Ricardo Monalban, see if i got it right. $\endgroup$
    – Will Jagy
    Apr 25, 2012 at 4:06
  • $\begingroup$ No, maybe it was first Star Trek Klingons. $\endgroup$
    – Will Jagy
    Apr 25, 2012 at 4:08
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    $\begingroup$ Well, I'm glad we cleared that up. $\endgroup$ Apr 25, 2012 at 4:24
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    $\begingroup$ @Gerry, in Miss Congeniality, Michael Caine gives instruction to people beautifying Sandra Bullock: "Eyebrows--there should be two!" en.wikipedia.org/wiki/Miss_Congeniality_%28film%29 $\endgroup$
    – Will Jagy
    Apr 25, 2012 at 4:37

The Chaz said:

""- use Alpern's solver - which seemed to indicate that there are no solutions (though I might have made an illegal substitution, so to speak)

I was able to identify $A = 6, B = 3, C = 6$ as solutions of $ \ 3A + 2B \ ^2 = C \ ^2$, but those aren't squares!""

You forgot that the useful and nice Alpern solver solves only in two variables; and here we have three variables.

I am surprised Gerry and Will did not correct you, Gerry is too busy correcting spelling or the aesthetics of the writing representation of mathematics; in particular not allowing a simpler and less "beautiful", but correct, ASCII representation of them. Very nice, i do certainly agree, the Dayo proof of no solutions to the equation. A short and simple and elegant proof at a time.

Post Scriptum: I do not dislike editing myself when i make a typographic error and maths have also to be well written and well expressed. They are ( Do maths have a plural in English like in Spanish and French ?) also a kind of literature where things have to be said well. But today i wrote an answer to somebody in which i said his question was also sometimes named the Polignac conjecture. Later someone edited me and gave a link to a wikipedia article on this conjecture. I like Wikipedia and can say i have learnt, as an amateur, a few good things with it. The matter and still slight problem is i did not put any link, somebody put it for me without my consent. What will be next, change the style i (we) use to write; the verbs i (we) use ? I mean there must be a limit of some sort to the editing mania.


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