Respected Mathematicians,

The Diophantine equation

$$2^x + 5^y = z^2$$

has solutions $$x = 3, y = 0, z = 3$$ and $$x = 2, y = 1, z = 3$$ I got these solutions by trial and error method. To be honest, these solutions are below the number $5$. So, I easily verified them by trial and error method. I would like to know the method which will give the solutions of the above equation, as well as the solutions of equations below.


$$4^x + 7^y = z^2$$


$$4^x + 11^y = z^2$$

Looking forward to your solution and support. baba


I'll do a piece of it to show you some methods you can try on the other pieces.

$2^x+5^y=z^2$. Let's do the case where $y=2s$ is even.

$2^x=z^2-(5^s)^2=(z+5^s)(z-5^s)$, so $z+5^s=2^m$ and $z-5^s=2^n$ with $m+n=x$. Eliminating $z$, $2\times5^s=2^m-2^n$, so $5^s=2^{m-1}-2^{n-1}$. The left side is odd, so the right side is odd, so $n=1$, and $5^s=2^{m-1}-1$. Left side is 1 modulo 4, so right side is 1 modulo 4, so we must have $m=2$. So if there's a solution with $y$ even, then $x=3$, $y=0$, $z=3$.

  • $\begingroup$ ! wonderful explanation you have given. Thank you. $\endgroup$ – baba Jan 21 '12 at 4:20

Assuming x, y, z are non-negative integers, there are no solutions to $4^x + 7^y = z^2$ and $4^x + 11^y = z^2$. This can be shown through modular arithmetic.

Taking the equation a) mod 3 you'll find the following:

  • $4^x \equiv 1 \pmod{3}$
  • $7^y \equiv 1 \pmod{3}$
  • $z^2 \equiv 0$ or $1 \pmod{3}$

Since $1 + 1 = 2$ and there are no squares equivalent to $2$ (mod 3), there are no solutions to $4^x + 7^y = z^2$.

Now taking equation b) mod 3:

  • $4^x \equiv 1 \pmod{3}$
  • $11^y \equiv (-1)^y \pmod{3}$
  • $z^2 \equiv 0$ or $1 \pmod{3}$

Since, once again, there are no squares equivalent to $2$ (mod 3), $y$ must be odd. However taking the equation mod 4 you'll get the following:

  • $4^x \equiv 0 \pmod{4}$
  • $11^y \equiv (-1)^y \pmod{4}$
  • $z^2 = 0$ or $1 \pmod{4}$

$y$ cannot be odd since this would imply that $11^y$ is equivalent to $-1 \equiv 3$ (mod 4) and there are no squares equivalent to 3 mod 4. This is a contradiction and therefore there are no solutions to equation b) either.

EDIT: Actually I just realised that I neglected the case where $x = 0$ and thus $4^0 \equiv 1 \pmod{4}$. But taking the resulting equation $1 + 11^y = z^2$ mod 5 proves that there's no solution anyway, as the quadratic residues for modulo 5 are 0, 1 and 4.

  • $\begingroup$ Curiously, $4^x+11^y=z^3$ has at least 2 solutions. $\endgroup$ – Gerry Myerson Jan 20 '12 at 12:19
  • $\begingroup$ @Gerry Myerson! can you show those two solutions please. $\endgroup$ – baba Jan 21 '12 at 4:19
  • $\begingroup$ $4^2+11^1=3^3$, $4^1+11^2=5^3$. $\endgroup$ – Gerry Myerson Jan 21 '12 at 5:18
  • $\begingroup$ @Gerry Myerson! Wow! great. I am searching solutions in above 10. Anyhow, thank you so much for this help. Could you suggest some equations like this and could you discuss the method for obtaining such solution without trail method please. $\endgroup$ – baba Jan 21 '12 at 7:14
  • $\begingroup$ Some people who have written papers on equations like these are L J Alex, L L Foster, and J L Brenner. The keyphrase is "exponential diophantine equations," although if you type that into Google, some of what comes up will be very technical and advanced work. $\endgroup$ – Gerry Myerson Jan 22 '12 at 1:55

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