# Find positive integer $x,y$ such that $7^{x}-3^{y}=4$

Find all positive integers $x,y$ such that $7^{x}-3^{y}=4$. It is the problem I think it can be solve using theory of congruency. But I can't process somebody please help me . Thank you

Let us go down the rabbit hole. Assume that there is a solution with $x, y > 1$, and rearrange to find

$$7(7^{x-1} - 1) = 3(3^{y-1} - 1)$$

Note that $7^{x-1} - 1$ is divisible by $3$ exactly once (since $x > 1$): the contradiction will arise from this.

Reducing modulo $7$ we find that $3^{y-1} \equiv 1$, and since the order of $3$ modulo $7$ is $6$, we find that $y-1$ is divisible by $6$, hence $3^{y-1} - 1$ is divisible by $3^6 - 1 = 2^3 \times 7 \times 13$. Now, reducing modulo $13$ we find that $7^{x-1} \equiv 1$, and since the order of $7$ modulo $13$ is $12$, we find that $x-1$ is divisible by $12$. As above, this implies that $7^{x-1} - 1$ is divisible by $7^{12} - 1$, which is divisible by $9$. This is the desired contradiction, hence there are no solutions with $x, y > 1$.

For an outline of the method I used here, see Will Jagy's answers to this related question.

• I did a new one yesterday by a different OP, that answer, while computationally difficult, does not have all the extra discussions of my earlier answers: math.stackexchange.com/questions/1946621/… Sep 30, 2016 at 16:44
• I don't understand why $a|b$ implies that $3^{a}-1|3^{b}-1$, and I wasn't able to prove it myself. Jan 8, 2021 at 6:29
• $3^{b}-1=3^{ak}-1=(3^{a}-1)(3^{a(k-1)}+3^{a(k-2)}+...+1)$? But then doesn't the parity of $b$ matter? Jan 8, 2021 at 6:35
• @DerekLuna Why should the parity of $b$ matter? Your identity proves the result by itself already. Jan 8, 2021 at 7:00
• I wasn't sure if everything in the middle canceled depending on the parity of $b$, but I see that it does now! Jan 8, 2021 at 7:01

If you take this equation mod 7 and mod 3, it becomes clearer:

$$7^x-3^y \equiv 4 \mod 7 \implies -3^y \equiv 4 \mod 7$$ $$7^x-3^y \equiv 4 \mod 3 \implies 7^x \equiv 4 \mod 3$$

This isolates the variables. You can either use Chinese Remainder Theorem or convert the new modular equations to normal equations to continue (as far as I can see).

• #BenLaurense: Please write the full solution . Sep 30, 2016 at 6:14
• Modular arithmetic won't carry you al the way: here is an obvious solutoin $x=y=1$ and we suspect this is the only one Sep 30, 2016 at 6:21