$$3x^2 + 2y^4 = z^4$$
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?
Best Answer
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.