Solve $7^a+b!=13^c$ over $\mathbb{Z}^{+}$

Question

Solve the following Diophantine equation $7^a+b!=13^c$ over positive integers.

Clearly $7^1+3!=13^1$ and $7^2+5!=13^2$. Are there any more solutions? Here are some thoughts.

$b!$ must be even, so $b>1$.

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Case 1: $c$ is even: let $c=2k$. Then, $b!\equiv13^c\equiv13^{2k}\equiv1\pmod 7\implies b=5$. So we're looking for solutions to $7^a+120=13^{2k}$. In modulo 4, this reduces to $(-1)^a\equiv1$ which means that $a$ is even: $a=2n$ and then we have $7^{2n}+120 = 13^{2k} \implies(13^k-7^n)(13^k+7^n)=120$. The possible even factors (they must both be even) are:

1. $2\cdot60 \implies 13^k=31, 7^n=29$;
2. $4\cdot30 \implies 13^k=17, 7^n=13$;
3. $6\cdot20 \implies 13^k=13, 7^n=7$;
4. $10\cdot12 \implies 13^k=11, 7^n=1$.

And clearly, only the third case yields a solution in integers with $k=n=1$.

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Case 2: $c$ is odd: let $c=2m+1$. Then, $b!\equiv13^c≡-1\pmod7 \implies b=3$ or $b=6$.

Sub-case: $b=6$. We have $7^a+720=13^{2m+1}$. In modulo 8 this becomes $(-1)a\equiv5^{2m+1}≡5$ which is not possible.

Sub-case: $b=3$. We have $7^a+6=13^{2m+1}$ [In modulo 13 this becomes $7^a+6≡0$ which means $a=12q+1$ and then…].

In modulo 4, this reduces to $(-1)^a+2\equiv1$ which means that $a$ is odd: $a=2p+1$ and then we have

$$ \begin{aligned} & 7^{2 p+1}+6=13^{2 m+1} \\ & \implies 7^{2 p+1}-7=13^{2 m+1}-13 \\ & \implies 7(7^{2 p}-1)=13(13^{2 m}-1) \\ & \implies 7(7^p-1)(7^p+1)=13(13^m-1)(13^m+1). \end{aligned} $$

I suspect $p=m=0$ is the only possible solution but cannot see a proof yet...

Answer 1

The Dipohatine equation

$7^a+6=13^c$

has no solutions when $a\ge2$. We prove this with some fairly creative modular arithmetic.

Modulo 49

If $a\ge 2$, then $7^a\overset{\text{forced}}{\equiv}0\bmod49$ and so $13^c\equiv6$. From calculating powers of $13\bmod 49$ we then infer that $c\equiv11\bmod14$. The modulus on the exponent is reduced to $14$ because $13^{14}\equiv1\bmod49$.

With the modulus on $c$ reduced to $14$, we seek a different modulus whose Euler totient is divisible by $14$.

Modulo 29

Since $c\equiv11\bmod14$ and $13\equiv10^2$ is a quadratic residue $\bmod29$, we infer that $13^c\equiv13^{11}\equiv4\bmod29$. But then $7^a\equiv27$ whereas $7\equiv6^2$ is a quadratic residue and $27\equiv3×3^2$ is not. (The number $29$ is $\not\equiv\pm1\bmod12$.) $\rightarrow\leftarrow$