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Solve the diophantine equation $3\,{a}^{3}b-13\,{b}^{3}-26\,a-24\,b=0.$ I have found two obvious solutions $a=b=0$ and $a=b=5.$ Are there another solutions?

InsideOut
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Leox
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  • Just a quick observation before I leave to catch the train ... $3b(a^3-2^3)=13(b^3+2a)$ and $b \ne 13$ so $b \mid 2a$ – rtybase Aug 09 '16 at 18:14
  • why $b \neq 0 \mod 13? $ – Leox Aug 09 '16 at 18:25
  • because if it was, https://www.wolframalpha.com/input/?i=3(x%5E3-8)%3D13%5E3+%2B+2x – rtybase Aug 09 '16 at 18:26
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    then what about $b=26, 39,\ldots ? $ – Leox Aug 09 '16 at 18:30
  • Yes, that observation didn't work. – rtybase Aug 09 '16 at 20:47
  • However, case 1: $a=b$ leads to https://www.wolframalpha.com/input/?i=3x%5E4-13x%5E3-26x-24x%3D0, i.e. $0, 5$ as the only solutions for this case. I will abuse Wolfram Alpha, but essentially, https://en.wikipedia.org/wiki/Rational_root_theorem is all we need. – rtybase Aug 09 '16 at 20:52
  • Case 2: $\gcd(a,b)=1$ leads to $b \cdot Q = 26 \cdot a$ which means $b=1$, $b=2$, $b=13$ or $b=26$ and again, none of the resulting polynomials in $a$ have integer solutions. E.g. $b=1$ yields $3a^3-26a-37=0$ and according to rational root theorem an integer root of this polynomial should divide 37, which is a prime and $a=1$ or $a=-1$ are not solutions. – rtybase Aug 09 '16 at 20:58
  • The remaining case is $\gcd(a,b)=k>1$ ... which, so far, looks tricky. – rtybase Aug 09 '16 at 20:59
  • Some context could be helpful. Where did this equation come from? – Yuriy S Aug 14 '16 at 23:29

1 Answers1

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Let us try to find a non-zero solutions of Diophantine equation.

Let $$c=\gcd(a,b),\quad a=cx,\quad b=cy,$$ then $$x(3c^3x^2y-26)=y(13c^2y^2+24),$$ $$y\,|\,(3c^3x^2y-26),\quad y\,|\, 26,$$ $$3c^3x^3-\dfrac{26}yx = 13c^2y^2 + 24,\quad y\in\{\pm1,\pm2,\pm13,\pm26\}.$$ Taking in account that for any values of $y$ the implicit function $x(c)$ is the asymptote on the coordinate axes and that negative values of $c$ correspond to the opposite values $y,$ we are at each value of $ y $ can get all integer solutions of this equation $(1)$.

All values $(x,y),$ which gives $c\gtrsim 5,$ obtained with using of Mathcad package, are the next:

enter image description here

With the graphic1 for $y=\pm1,$ graphic2 for $y=\pm2,$ graphic3 for $y=\pm13$ and graphic4 for $y=\pm26$ we have all possible integer roots for $x,y,c$.

Among them only $c(1,1) = 5$ is integer. Substituting of $a=b=5$ to the original equation shows that this is the right solution.

So $$\boxed{(a=0,b=0) \vee (a=5,b=5)}$$ are the all integer solutions of the given Diofantine equation.

Note

Of course, in each of these cases, we can show the presence or absence of integer roots using the Rational Root Theorem. However, in this case that looks like a very technical and overload the main logic of proof.

Yuri Negometyanov
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