Can anyone please tell me a general method to solve non-linear Diophantine equation like

$8^x + 15^y=17^z$

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

  • $\begingroup$ Welcome to Math SE. There's no general method I know of to solve non-linear Diophantine equations. However, note the Beal conjecture says that since $8,15,17$ are coprime, there's no solutions for the first one for $x,y,z \gt 2$. If so, you only need to check for $0 \le x,y,z \le 2$ (with only $x=y=z=2$ working in this case, i.e., $8,15,17$ is a Pythagorean triple). $\endgroup$ – John Omielan Oct 17 '19 at 6:42
  • $\begingroup$ Gave the wrong link earlier – here's the link for the 1st equation: math.stackexchange.com/questions/2494848/… $\endgroup$ – Gerry Myerson Oct 17 '19 at 11:18

As others have pointed out , there is no general method to solve such equations .

If you want a solution of the two equations you asked , they are as follows:

$\color{blue}{1.(8^x + 15^y = 17^z)}$

  • Show that $8,15,17$ are co-prime and then use Beal Conjecture to show that no solution can exist for $x,y,z \gt2.$

    It is easy to conclude that the only solution is $x=y=z=2.$

$\color{red}{2.(x^3 + 2y^3 = 4z^3)}$

  • It is easy to verify that $x \, ,y \, , z = 0 $ satisfies the relation . All that is left is to prove that there is no more solutions in positive integers.

    Let us assume $(a,b,c)$ to be the smallest value which satisfies the relation. Now since $$a^3 + 2b^3 = 4c^3$$ We conclude that $a$ must be divisible by $2.$ Let us further assume that $a=2k$ , then putting this value back into the equation , we get : $$8k^3 + 2b^3 = 4c^3$$ which reduce to $$4k^3 + b^3 = 2c^3$$

Again we conclude that $b$ must also be divisible by $2.$ Similarly by putting $b=2l$ , reduce the equation and again conclude that $c$ must also be divisible by $2.$ Again $c=2m$ , we get the following relation : $$k^3+2l^3 = 4m^3$$ Hence we have found a smaller triplet $(k,l,m)$ , which satisfies the relation .
But this a contradiction to the fact that $(a,b,c)$ are the smallest triplet.

So , we can conclude that there is no solution other than $(0,0,0)$ which fulfills the condition.

  • 2
    $\begingroup$ You can't use the Beal conjecture to prove anything, as it's a conjecture, not a theorem. $\endgroup$ – Gerry Myerson Oct 17 '19 at 11:19

If you want a general algorithm, you're bound to be disappointed since you're asking for way too much. No such general method exists even when you restrict to polynomial equations, and studying special families of nonlinear equations are the subject of much of Number Theory!

To list but a few examples of extremely hard nonlinear problems: Fermat's Last Theorem with the equation $x^n+y^n=z^n$ for $n\geq3$), Catalan's Conjecture (now Mihailescu's theorem) with equation $x^a-y^b=1$, and $x^3+y^3+z^3=n$ for fixed $n$ (see the Wikipedia page on sum of three cubes).

In fact, that "no such general method exists" can even be made precise. Hilbert's $10$th problem asks whether or not there is a general algorithm to determine whether or not a given Diophantine equation has integer solutions. This problem has now been resolved by Matiyasevich, building on the work of others, proving that no such general algorithm exists, even in theory.


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