Diophantine equation power of 7 and 2
Solution 1:
Let $x>2$ and $y>4$.
Rewrite our equation in the following form: $$49(7^{x-2}-1)=48(2^{y-4}-1),$$ which says that $2^{y-4}-1$ is divisible by $49$,
which says that $y-4$ is divisible by $21,$ which says $2^{y-4}-1$ is divisible by $2^{21}-1=49\cdot127\cdot337,$
which gives that $7^{x-2}-1$ is divisible by $337$,
which says $x-2$ is divisible by $56$ (thanks to dear Will Jagy).
and from here $7^{x-2}-1$ is divisible by $7^{56}-1=2^6\cdot3\cdot5^2\cdot29\cdot113...,$
which gives $48(2^{y-4}-1)$ is divisible by $64$, which is a contradiction.
Id est, our equation has no natural solutions for $x>2$ and $y>4$.
Can you end it now?
Solution 2:
CW answer, votes don't affect me for this one.
There is a very good method for
$$ a p^m = b q^n + c, $$ where all are positive integers and $p,q$ are prime
discovered by https://math.stackexchange.com/users/292972/gyumin-roh
Exponential Diophantine equation $7^y + 2 = 3^x$
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
Hello, Sailor
There was a girl in high school, active in "forensics" which was combined debate and related competition among many schools. She had practiced a really excellent Hello, Sailor. At the time, about 1974...
Eric Idle wrote Hello Sailor, his first novel, in 1970
A book of the same title was mentioned by Idle and Cleese in the Monty Python's Flying Circus episode "Sex and Violence" during "The Wrestling Epilogue" sketch, in which a humanist philosophy professor, author of a novel entitled "Hello Sailor," debates an Anglican monsignor over the existence of God in an officiated wrestling match.
https://en.wikipedia.org/wiki/Hello,_sailor
should bump question in active queue ... appears MIchael's answer does appear first maybe for being accepted .. Seems appropriate ... compare active queue after deleting
Solution 3:
I edit my previous answer. My only purpose here is to give an answer distinct from that given by the distinguished friend Michael Rozenberg.
We easily verify that $y=1$ and $y=4$ give two solutions and that $y=2$ and $y=3$ must be discarded; also $x$ must be even (reducing modulo $16$) so we consider the new equation $$7^{2x}=3\cdot2^{4+y}+1\iff(49)^x=48\cdot2^y+1;\space x\ge1, \space y\ge1$$ Now if $x$ is even then $$1\equiv8\cdot2^y+1\pmod{10}\Rightarrow 0\equiv2^{y+3}\pmod{10}$$ which is not possible so $x$ should be odd.
On the other hand we have $$(48+1)^x=48^2M+48x+1=48\cdot2^y+1\Rightarrow48M+x=2^y$$ and $x$ should be even.
Since $x$ cannot be odd and even,the only solutions of the proposed equation are $(x,y)=(1,1),(2,4)$