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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

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2 Answers 2

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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.

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    $\begingroup$ 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/… $\endgroup$
    – Will Jagy
    Commented Sep 30, 2016 at 16:44
  • $\begingroup$ I don't understand why $a|b$ implies that $3^{a}-1|3^{b}-1$, and I wasn't able to prove it myself. $\endgroup$
    – Derek Luna
    Commented Jan 8, 2021 at 6:29
  • $\begingroup$ $3^{b}-1=3^{ak}-1=(3^{a}-1)(3^{a(k-1)}+3^{a(k-2)}+...+1)$? But then doesn't the parity of $b$ matter? $\endgroup$
    – Derek Luna
    Commented Jan 8, 2021 at 6:35
  • $\begingroup$ @DerekLuna Why should the parity of $ b $ matter? Your identity proves the result by itself already. $\endgroup$
    – Ege Erdil
    Commented Jan 8, 2021 at 7:00
  • $\begingroup$ I wasn't sure if everything in the middle canceled depending on the parity of $b$, but I see that it does now! $\endgroup$
    – Derek Luna
    Commented Jan 8, 2021 at 7:01
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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).

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  • $\begingroup$ #BenLaurense: Please write the full solution . $\endgroup$ Commented Sep 30, 2016 at 6:14
  • $\begingroup$ 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 $\endgroup$ Commented Sep 30, 2016 at 6:21

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