Solutions of some Diophantine equations

Question

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.

a)

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

b)

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

Looking forward to your solution and support. baba

Answer 1

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 $x=3$, $y=0$, $z=3$.

Answer 2

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.