Solve in positive integers: $5^x 7^y +4=3^z$

Question

Solve in positive integers: $5^x 7^y +4=3^z$.

I tried to solve it with log but I couldn't complete.

Answer 1

Rewrite this as

$$3^z-4=5^x7^y$$

Then we see, because $x>0$, that

$$3^z-4\equiv 0\mod 5 \iff 3^z\equiv -1\mod 5$$

so $z=2j$ is even.

Then we have $(3^j-2)(3^j+2)=5^x7^y$. Since the gcd of these two divides $4$--because $(3^j-2)+(3^j+2)=4$--we have that they are coprime (any larger factor would then be a factor of $5^x7^y$ which is odd), i.e. one of the following two cases hold

$$\begin{cases} 3^j+2=5^x\quad ,\quad 3^j-2=7^y \\ 3^j+2=7^y\quad ,\quad3^j-2=5^x\end{cases}.$$

In the second case we get $7^y\equiv 2\mod 3$ which is impossible, so we are in case $1$. Then $3^j=7^y+2$ so $j\equiv 2\mod 6$ which recovers our original conclusion, that $z=12\ell+4\equiv 0\mod 4$, which is a contradiction as this implies

$$-3\equiv 3^z-4\equiv 5^x7^y\equiv 0\mod 5.$$

Answer 2

[Based on the work from @Adam Hughes]

We assume $x>1$ and rewrite it:

$$3^z-4=5^x7^y \Rightarrow 3^z-4\equiv 0\mod 5 \iff 3^z\equiv -1\mod 5$$

so $z=2j$ is even.

Then we have $(3^j-2)(3^j+2)=5^x7^y$. Since the gcd of these two divides $4$--because $(3^j+2)-(3^j-2)=4$--we have that they are coprime (any larger factor would then be a factor of $5^x7^y$ which is odd), i.e. one of the following two cases hold

$$\begin{cases} 3^j+2=5^x\quad ,\quad 3^j-2=7^y \\ 3^j+2=7^y\quad ,\quad3^j-2=5^x\end{cases}.$$

Case 2: $3^j+2=7^y\quad ,\quad3^j-2=5^x \Rightarrow 7^y = 4 + 5^x$, which is impossible if you look at it $\mod 3$.

So we have Case 1: $3^j-2=7^y\quad ,\quad3^j+2=5^x \Rightarrow 7^y+4 = 5^x$. If you look at it $\mod 4$, you get $2|y$, so we write $y=2 y_2$

We look at it $\mod 25$: this gives us $7^y + 4 \equiv 49^{y_2} +4 \equiv (-1)^{y_2} + 4 \equiv 5^x \equiv 0$, which is impossible.

it remains to look at x=1, here we have the only solution $y=0,x=1,z=2$ (you can ignore this if you only want strictly positive solutions)

Answer 3

$5^x7^y+4=3^z$ with $x,y,z\in\Bbb Z^+$.

mod $5$ gives $z=2z_1,\, z_1\in\Bbb Z^+$. Then $5^x7^y=\left(3^{z_1}+2\right)\left(3^{z_1}-2\right)$.

$\gcd(3^{z_1}+2, 3^{z_1}-2)=1$, so two cases:

• $7^y=3^{z_1}+2$, impossible mod $3$.
• $5^x=3^{z_1}+2$. mod $3$: $x$ is odd. mod $8$: $z_1$ is odd.
  $7^y=3^{z_1}-2$. mod $7$: $z_1$ is even. Contradiction.