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This question: "Solve: $a!+5^b=7^c$" was closed for lack of context, and probably lack of effort by OP. Nonetheless, the question is interesting in itself, and several particular solutions were given in the comments. I show here those given answers are nearly exhaustive, and pose a question about the one instance for which I have not yet obtained a which a full solution.

  1. Based on parity alone, $a!$ must be even, hence $a \ge 2$

  2. $\min (a!+5^b)=2!+5^0=3$, hence $c \ne 0$

  3. If $a\ge 7$, then $7\mid a! \land 7\mid 7^c \Rightarrow 7\mid 5^b$. This is not possible, so $a<7$

  4. Examine the case $b=0$. Here, $2\le a \le 6 \Rightarrow a! \in \{2,6,24,120,720\}$ and $a!+5^0=a!+1 \in \{3,7,25,121,721\}$. The only instance in which $a!+5^0=7^c$ is $3!+5^0=7$. Solution 1 is $a=3,b=0,c=1$

  5. If $b>0 \land a\ge 5$, then $5\mid a! \land 5\mid 5^b \Rightarrow 5\mid 7^c$. This is not possible, so $a<5$

  6. $2\le a \le 4 \Rightarrow a! \in \{2,6,24\}$. We can examine each of these three cases individually.

  7. $24+5^b=7^c$ is analyzed modulo $7$ and modulo $5$. First, $3+5^b\equiv 0 \bmod 7$ is true when $b$ has the form $6m+2$. Next, $4\equiv 7^c \equiv 2^c \bmod 5$ is true when $c$ has the form $4n+2$. In order to have integer solutions, this equation must be formulated $24+5^{6m+2}=7^{4n+2}$. Rearranging, $24=7^{2(2n+1)}-5^{2(3m+1)}$. This is simply the difference of two squares, viz: $24=(7^{(2n+1)}-5^{(3m+1)})(7^{(2n+1)}+5^{(3m+1)})$. This resolves $24$ into two factors, both of which are even and have no common factors other than $2$. The possiblities for this are $24=2\times 12$ and $24=4\times 6$. Moreover, for $m>1$ or $n>1$, the factor $(7^{(2n+1)}+5^{(3m+1)})$ is much larger than $24$ itself, so the only possible candidate is given by $m=n=0$, which sets $24=(7-5)(7+5)=2\cdot 12$ which is true, leading uniquely to Solution 2: $a=4,b=2,c=2$

  8. $6+5^b=7^c$ is analyzed modulo $7$ and modulo $5$. First, $6+5^b\equiv 0 \bmod 7$ is true when $b$ has the form $6m$. Next, $1\equiv 7^c \equiv 2^c \bmod 5$ is true when $c$ has the form $4n$. As before, we derive $6=(7^{(2n)}-5^{(3m)})(7^{(2n)}+5^{(3m)})$. This plainly has no solutions because each of the factors is even and contains at least one factor of $2$, but $6$ has only one factor of $2$. So there are no solutions in this instance.

  9. $2+5^b=7^c$ is analyzed modulo $7$ and modulo $5$. First, $2+5^b\equiv 0 \bmod 7$ is true when $b$ has the form $6m+1$. Next, $2\equiv 7^c \equiv 2^c \bmod 5$ is true when $c$ has the form $4n+1$. In order to have integer solutions, this equation must be formulated $2+5^{6m+1}=7^{4n+1}$. By inspection, when $m=n=0$, we get Solution 3: $a=2,b=1,c=1$. Here I am stopped. Since a solution exists, it is not likely that I can rule out larger solutions (i.e. for $mn \ge 1$) by modular arithmetic, and since the exponents in this case are odd, I cannot use the difference of two squares trick which afforded solutions in the previous cases.

My question is: Is there a method or strategy for evaluating the existence and nature of solutions to $2+5^{6m+1}=7^{4n+1}$ for $mn \ge 1$?

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    $\begingroup$ I believe (but am not certain) that $n=2$ is one of the known cases of the generalized Catalan Conjecture. Perhaps the references in that link would settle the issue. $\endgroup$
    – lulu
    Commented Oct 15, 2021 at 17:20

3 Answers 3

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If $mn\geqslant 1$, then one has $$5^{6m+1}\equiv 25\pmod{100}\qquad \text{and}\qquad 7^{4n+1}\equiv 7\pmod{100}$$ So, if $mn\geqslant 1$, then $2+5^{6m+1}=7^{4n+1}$ has no solutions.

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  • $\begingroup$ Great answer. Thanks! $\endgroup$ Commented Oct 15, 2021 at 17:52
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Actually, as to your paragraph 9,

Since a solution exists, it is not likely that I can rule out larger solutions

is too pessimistic; a string of modular statements may work. Your $7^m = 5^n + 2$ becomes $7^m -7 = 5^n -5$ or $$ 7 (7^x - 1) = 5 ( 5^y - 1) $$ We assume $x,y \geq 1$ and hunt for a contradiction.

(I) since $7^x \equiv 1 \pmod 5,$ we calculate $4 | x.$ Well $ 7^4 - 1 = 96 \cdot 25.$ As $4|x,$ we know $7^x - 1$ is divisible by said $2400;$

(II) $7^x - 1$ is divisible by $25 = 5^2,$ but $5(5^y-1)$ is not divisible by $25$ Therefore $ 7 (7^x - 1) \neq 5 ( 5^y - 1) $

unless $x,y=0;$ there is no larger solution.

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

here are some links on this method, including the guy from whom I learned it

https://math.stackexchange.com/users/292972/gyumin-roh

Exponential Diophantine equation $7^y + 2 = 3^x$ Gyumin Roh answer from 2015

Elementary solution of exponential Diophantine equation $2^x - 3^y = 7$.

Elementary solution of exponential Diophantine equation $2^x - 3^y = 7$. ME! 41, 31, 241, 17

Finding solutions to the diophantine equation $7^a=3^b+100$ 343 - 243 = 100

http://math.stackexchange.com/questions/2100780/is-2m-1-ever-a-power-of-3-for-m-3/2100847#2100847

The diophantine equation $5\times 2^{x-4}=3^y-1$

Equation in integers $7^x-3^y=4$

Solve in $\mathbb N^{2}$ the following equation : $5^{2x}-3\cdot2^{2y}+5^{x}2^{y-1}-2^{y-1}-2\cdot5^{x}+1=0$

Solve Diophantine equation: $2^x=5^y+3$ for non-negative integers $x,y$. 128 - 125 = 3

Diophantine equation power of 7 and 2

Find natural numbers a,b such that $|3^a-2^b|=1$ did +-1

Finding all natural $x$, $y$, $z$ satisfying $7^x+1=3^y+5^z$

https://math.stackexchange.com/questions/4155854/how-do-we-find-all-k-s-in-mathbbn-where-5s-2-3k/4156003#4156003 5-3=2

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

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  • $\begingroup$ Very good references! I can see now that mathlove's answer was not just a bit of arcane knowledge on his part, but can be logically arrived at. Thanks. $\endgroup$ Commented Oct 16, 2021 at 15:03
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Once you establish $2\le a\le4$, you can treat the six possibilities with $b\lt2$ individually, so it remains to consider $a!+5^b=7^c$ with $b\ge2$, in which case we have $7^c$ mod $25$ is either $2$, $6$, or $-1$. But the powers of $7$ mod $25$ are $1$, $7$, $-1$, and $-7$. So the only possibility with $b\ge2$ is $4!+5^b=7^c$. Moreover, closer inspection of the powers of $7$ mod $25$ shows $7^c\equiv-1$ mod $25$ implies $c$ is even (and positive), at which point an argument mod $8$ tells us $b$ must also be even. Since $7^{2m}-5^{2n}=(7^n-5^n)(7^n+5^n)\ge24$ if $m$ and $n$ are both positive, we find $4!+5^2=7^2$ is the only solution with $b\ge2$.

(For completeness, of the six possibilities with $b\lt2$, only two are powers of $7$, namely $3!+5^0$ and $2!+5^1$.)

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