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Find all solutions of $9^x-4^y=5^z$ over the naturals.

My attempt: We have

$${(3^x)}^2-{(2^y)}^2=5^z \Rightarrow (3^x+2^y)(3^x-2^y)=5^z.$$

It can be easily proved that $\gcd (3^x+2^y,3^x-2^y)=1$ so we have

$$3^x-2^y=1,3^x+2^y=5^z.$$

What do we have to do from here?

Xam
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Taha Akbari
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    You can solve the equation $3^x-2^y=1$ using modular arithmetic. – Xam Jul 10 '17 at 17:45
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    Levi ben Gershon (Gersonides) theorem on "harmonic numbers". https://math.stackexchange.com/questions/164874/2x-3y-1-has-only-three-natural-pairs-as-solutions – Robert Israel Jul 10 '17 at 17:50
  • @Xam modular what?the equation $2^y-3^x=1$ can be solved that way but I am not sure about this. – Taha Akbari Jul 10 '17 at 17:50
  • @TahaAkbari I mean, using congruences. See the link given by Robert Israel. – Xam Jul 10 '17 at 17:53

1 Answers1

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As @ Xam said,

if $y=0$, then there is no solution,

if $y=1$, then $x=1$.

Now if $y$ is greater than $1$, then $2^y$ is divisible by $4$. using modular arithmetic module $4$ we have:

$(-1)^x-0$ is congruent to $1$ module $4$, so $x$ must be even, i.e. $x=2k$.

Now $2^y=3^{2k}-1=(3^k-1)(3^k+1)$, and by the same conclusion you have done, as you explained, we can conclude that:

$(3^k-1)=2$ and $(3^k+1)=2^{y-1}$,

so $k=1$ and $x=2$ and $y-1=2$.

But this values for $x$ and $y$ does not give any solution to the main equation.

Another soloution:

We claim that $y$ could not be greater than $1$.

Suppose on contrary, that y is greater than $1$, then $4^y$ is divisible by $8$. Using modular arithmetic module $8$ we have:

$1-0$ is congruent to $5^z$ module $8$, so $z$ must be even, i.e. $z=2t$.

Again by considering the main equation module $3$, we have :

$0-1$ is congruent to $(-1^2)^t$ module $3$, i.e. $-1=1$ module $3$, which is a contradiction.

So we have two cases:

  • $y=1$, then $(3^x-2)(3^x+2)=5^z$, and by the same conclusion as you have been used, we can conclude that:
    $(3^x-2)=1$ and $(3^x+2)=5^z$, which gives:
    $x=1$ and $z=1$.

  • $y=0$, then $(3^x-1)(3^x+1)=5^z$, we can conclude that:
    $(3^x-1)=2$ and $(3^x+1)=(1/2)5^z$, which does'nt gives any solution.

Davood
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