Answer: $(3,0,1),(2,1,1),(4,2,2)$ $a=0,1$ gives no solution by Catalan's conjecture. $b=0$ gives $(a,b,c)=(3,0,1)$ by $\pmod{7}$ and a very little casework. $c=0$ gives no solution because it yields $a!+5^b=1$. Now, assume that $a,b,c$ are positive integers. By $\pmod{5}$ and Catalan's conjecture, $a \in \{2,3,4\}$. 1. $a=2$ $\pmod{3}$ shows $b=2x+1$ is odd. $\pmod{4}$ shows $c=2y+1$ is odd. Assume $x,y\ge 1$. Rewritting the question and taking $\pmod{25}$ gives $$2+5\cdot25^x = 7\cdot 49^y$$$$2\equiv 7(-1)^y\pmod{25} $$So either of $x$ or $y$ is $0$. This gives $(2,1,1)$. 2. $a=3$ Do the same as above (Take $\pmod{25}$). However, it yields no solution. 3. $a=4$ Do the same as above (Take $\pmod{25}$). It gives $(4,2,2)$.

a!+5^b=7^c if a>=5... a!+5^b mod5== 0 then a<5if a=4......... 24+5^b=7^c 5^b+24 mod4==1 Then 7^c mod4==1 then c=2k and 7^c mod3==1 Then 24+5^b mod3==1 then b=2s Now (7^s_5^b)(7^s+5^b)=24 We arrive at these answers by examining the cases. (s,b)=(2,2) if a=3 6+5^b=7^c. 7^c mod3==1 then 6+5^b mod3==1 as a result b=2k...... 6+5^b mod25==6 But 7^c mod25== 7,24,18,1 then a=3 there is no answer. a=2........ 2+5^b=7^c........ 7^c mod3==1 Then 2+5^b mod3==1 as the result b=2k+1 if b>2 2+5^b mod25== 2 but 7^c mod25=7,24,18,1 There is no answer.. But b=1 and c=1 in this case Ture.. if a=1 5^b+1=7^c 7^c mod6==1 but 1+5^b mod6==0,2 So contradiction ، there is no answer. (a,b,c)=(2,1,1) and (4,2,2)

$Case$ $I:$ $b=0$: Then $a!+1=7^{c}$ and as $7|a!$ when $a\geq 7$(in that case $a!+1>0$, so $c>0$), then $a\leq 6$. A quick check gives the solution $(a,b,c)=(3,0,1)$. $Case$ $II:$ $b>0$. Then $a!+5^{b}>5^{b}>5$, so $c>0$. If $a>4$, then $5$ divides the LHS, but not the RHS, contradiction. Also, as $5^{b}\equiv 7^{c}\equiv 1$ (mod $2$), then $2$ divides $a!$, so $2\leq a\leq 4$. $Case$ $2.1$) $a=2$: $2+5^{b}=7^{c}$. If $b=1$ we reach the solution $(a,b,c)=(2,1,1)$. If $b>1$ modulo $25$ yields $7^{c}\equiv 2$ (mod $25$), which is a contradiction as $7^{c}\equiv 1,7,24,18$ (mod $25$). $Case$ $2.2$) $a=3$: $6+5^{b}=7^{c}$. Modulo $3$ gives that $b$ is even. Now looking modulo $25$, $LHS$ has residue $6$ and $7^{c}\equiv 1,7,24,18$ (modulo $25$), so there are no solutions in this case as $b>0$. $Case$ $2.3$) $a=4$: $24+5^{b}=7^{c}$. Again, modulo $3$ gives that $b$ is even. Modulo $25$ yields that $c$ is even, so $7^{2c_{1}}-5^{2b_{1}}=24$. However, $7^{c}-5^{b}\geq 7^{2c_{1}}-(7^{c_{1}}-2)^{2}=4\times (7^{c_{1}}-1)\geq 24$. Equality occurs iff $c_{1}=1$, so we reach the solution $(a,b,c)=(4,2,2)$. $\boxed{\text{Answer: }(a,b,c)=(3,0,1), (2,1,1), (4,2,2)}$

Afo wrote: Answer: $(3,0,1),(2,1,1),(4,2,2)$ $a=0,1$ gives no solution by Catalan's conjecture. $b=0$ gives $(a,b,c)=(3,0,1)$ by $\pmod{7}$ and a very little casework. $c=0$ gives no solution because it yields $a!+5^b=1$. Now, assume that $a,b,c$ are positive integers. By $\pmod{5}$ and Catalan's conjecture, $a \in \{2,3,4\}$. 1. $a=2$ $\pmod{3}$ shows $b=2x+1$ is odd. $\pmod{4}$ shows $c=2y+1$ is odd. Assume $x,y\ge 1$. Rewritting the question and taking $\pmod{25}$ gives $$2+5\cdot25^x = 7\cdot 49^y$$$$2\equiv 7(-1)^y\pmod{25} $$So either of $x$ or $y$ is $0$. This gives $(2,1,1)$. 2. $a=3$ Do the same as above (Take $\pmod{25}$). However, it yields no solution. 3. $a=4$ Do the same as above (Take $\pmod{25}$). It gives $(4,2,2)$. You didn't quite complete the third case as $24$ is a residue of $7^{c}$ mod $25$ and yields the solution $(a,b,c)=(4,2,2)$, but mod $25$ only gives $c\equiv 2$ (mod $4$), which can then be bounded, see my solution above.

Functional_equation wrote:

How can you get $ 7^c\equiv 1\equiv 2+5^b\equiv 2+(-1)^b\pmod 4\implies (-1)^b\equiv -1\pmod 4 $ in case 3.2?

AK001 wrote: Functional_equation wrote:

How can you get $ 7^c\equiv 1\equiv 2+5^b\equiv 2+(-1)^b\pmod 4\implies (-1)^b\equiv -1\pmod 4 $ in case 3.2? There should be (mod 6), not (mod 4) , I guess

zlatan.ibrahimovic84 wrote: $a!+5^b=7^c$ if $a\ge 5$... $a!+5^b \equiv 0 \pmod 5$ then $a<5$ if $a=4 \Rightarrow$ $24+5^b=7^c$ $ 5^b+24 \equiv 1\pmod4$ Then $7^c \equiv 1 \pmod4$ then $c=2k$ and $7^c \equiv 1 \pmod3$ Then $24+5^b \equiv 1 \pmod3$ then $b=2s$ Now $(7^s-5^b)(7^s+5^b)=24$ We arrive at these answers by examining the cases. $(s,b)=(2,2)$ if $a=3$ $6+5^b=7^c$. $7^c \equiv 1 \pmod3$ then $6+5^b \equiv 1 \pmod3$ as a result $b=2k\Rightarrow$ $6+5^b\equiv 6 \pmod{25}$ But $7^c\equiv 7,24,18,1 \pmod{25}$ then $a=3$ there is no answer. $a=2\Rightarrow$ $2+5^b=7^c \Rightarrow$ $7^c \equiv 1 \pmod3$ Then $2+5^b\equiv 1 \pmod3$ as the result $b=2k+1$ if $b>2 \Rightarrow$ $2+5^b\equiv 2 \pmod{25}$ but $7^c\equiv 7,24,18,1 \pmod{25}$ There is no answer.. But $b=1$ and $c=1$ in this case Ture.. if $a=1$ $5^b+1=7^c$ $7^c\equiv 1 \pmod6$ but $1+5^b\equiv 0,2 \pmod6$ So contradiction ، there is no answer. $(a,b,c)=(2,1,1)$ and $(4,2,2)$

@Marinchoo for your third cases, after you got b and c is even. just use the a^2 - b^2 = (a+b)(a-b) , there is no need to use some complex equation and what it is.

If $a \ge 7$ then we have no solutions. So $a \le 6.$ If $a=0$ or $a=1$: By catalan's conjecture, no solutions. If $a=2$: $5^b+2=7^c$, the only solution in this case is $(2,1,1)$ If $a=3$: $5^b+6=7^c$ no solution in this case If $a=4$: $5^b+24=7^c$, only solution is $(4,2,2)$ If $a=5$: $5^b+120=7^c$, no solution. If $a=6$: $5^b+720=7^c$, no solution. so the only solutions are: $\boxed{(4,2,2) or (2,1,1)}.$

For $a\geq5$,since both $a!$ and $5^{b}$ is divisible by 5, we got $7^{c}\equiv 0\mod5$,which is not possible For $a=4$, we got $24+5^{b}=7^{c}$. $b=0$ is trivial. For $b\geq1$, we got $7^{c}\equiv 4\mod5\Rightarrow c=4k+2=2(2k+1)\hspace{4mm}k\in \mathbb{N}$. $c=0$ is also trivial. We can see that $5^{b}\equiv 4\mod7\Rightarrow b=6m+2=2(3m+1)\hspace{4mm}m\in \mathbb{N}$ $$(7^{2k+1})^2-(5^{3m+1})^2=24$$$$(7^{2k+1}+5^{3m+1})(7^{2k+1}-5^{3m+1})=24\Rightarrow k=0\hspace{3mm}m=0\Rightarrow b=2\hspace{3mm}c=2$$ For $a=3$, we got $5^{b}+6=7^c$ From $b=0$, we got $c=1$. $c=0$ is trivial. For $b\geq1$ and $c\geq1$ $$7^{c}\equiv1\mod5\Rightarrow c=4k=2(2k)\hspace{2mm}k\in\mathbb{Z}^+$$$$5^{b}\equiv1\mod7\Rightarrow b=6m=2(3m)\hspace{2mm}m\in\mathbb{Z}^+$$$$(7^{2k})^2-(5^{3m})^2=6$$Which is not possible. For $a=2$, we got $5^b+2=7^c$. $b=0$ and $c=0$ is trivial. For $b=1$, we got $c=1$. For $b\geq2$: $$5^b+2\equiv2\mod25\Rightarrow 7^c\equiv2\mod25$$Which is not possible. And for both $a=1$ and $a=0$,we got: $7^c-5^b=1$ which is not true prior to the Catalan Conjecture

My Solution:Lets a>=5 then a!+5^b=5 mod 10 but it impossible for 7^c then a<5 here we have 4 cases 1)a=0 no solution because 5^b+1 is 6 mod 10 but its impossible for 7^c give 6 mod 10 2) Lets a=1 then it is similar like a=0 no solution 3)a=2 here we can see that only solution is (a,b,c)=(2,1,1) Because 2+5^a=7^c no solution for a,c>1 at mod 10 4) Lets a=3 then 6+5^b=7^c Can have solution if c=4k but then 6+5^b=2401^k and looking at mod 4 LHS give 3 but RHS 1 Then no solution for a=3 (but we can see)that for b=0 here is a solution (a,b,c)=(3,0,1) 5)last one is a=4 here >24+5^b=7^c here we can see that for (a,b,c)=(4,2,2) is a solution then lets prove that it's only solution for a=4.Here looking at mod 5 we can see that 7^c=4 mod 5 then c=4k+2 or 2(2k+1) you can see it looking at mod 5 for remainder 4 for powers of 7.Now let look at mod 7 here 5^b=4 mod 7 then b=6m+2 or 2(3m+1).And from 24=7^b-5^b we can write it as (7^(2k+1))²-(5^(3m+1))²=24 opening it bu formula we can give values gor both sides then we will get that m=0 and k=0 then b=2 c=2 and we proved that it is only solution at a=4 then Our final answer is (a,b,c)=(2,1,1).(3,0,1).(4,2,2) Then equation have 3 triples (a,b,c) solutions and we are done ))) (Notice that k and m are none negative integers)

note (b-1)^2 + (c-1)^2 neq 0 so if b > 1 then 5 not divide a! and otherwise c > 1 and 7 not divide a! thus a leq 6 so we take cases on a. if a=0 or a=1 (same case) we have 5^b+1=7^c and this is baddd by xig cuz its obv bad for b=0,1,2,3 and then since 7 divide 5^3+1 for b > 3 it must have primitive prime factor by xig so it doesnt werk if a=2 then note that b=1 works frmo c=1 and 0 doesnt and then otherwise lhs is always 2mod25 which is bad cuz powers of 7 are only 7 -1 -7 1 mod 25 now if a=3 if b=0 then c=1 and it werks otherwise since b is even (mod 3) lhs is 6 mod 25, absurd now if a=4 then its ez cuz u get by mod 3 that b even and then by mod 4 that c is even and get that (7^(c/2) - 5^(b/2)) (7^(c/2) + 5^(b/2)) = 24 and by size b/2,c/2 are both at most 1 so they have to be equal to 1 or 0 but c/2 =0 bad because it means 7^(c/2) - 5^(b/2) leq 0 so c=2 and checking b/2 =0 gives its bad so b=c=2 holds which works now onto the hardest cases where a=5 or a=6 here we have to take mod 5 and get that 5^b cant be 0 mod 5 so b =. 0 whichis bad as 121 721 arent power of 5 thus we are done and answers are (2,1,1) (3,0,1) (4,2,2)