PROBLEM
Solve the equation in the set of natural numbers:
$11^x=5*2^y+1$
WHAT I THOUGHT OF
We can write $11^x$ as $(10+1)^x$
We know that $(a+b)^x= M_a+b^x=M_b+a^x$
Applying the formula above we can write
$(10+1)^x=M_1+10^x$
$M_1$ basicly means every number so we are going to note it as $a$.
$a+10^x=10*2^{y-1}$
I dont know what to do further. This are all of my ideas. Hope one of you can help me!
Also, i can definately see that a solution is $(1,1)$, but i want a demonstration that shows this.
Number theory arguments seem to work best here.
First, notice that $y$ cannot be zero because we would get $11^x=6$.
Case 1: $y=1$. We obtain the solution $(x,y)=(1,1)$ in this case.
Case 2: $y\geq 2$. Observe that $x$ must be strictly positive. Now we work $\bmod{4}$. Notice that $2^y$ is a multiple of $4$, so $$11^x\equiv 5\cdot 2^y+1\pmod{4}$$ can be written as $(-1)^x\equiv 1\pmod{4}$. Hence, $x$ must be even. Write $x=2s$, and notice that in this case the equation can be written as $$(11^s-1)(11^s+1)=5\cdot 2^y.$$ Hence, the two factors will have the form $2^k$ and $5\cdot 2^l$ (not necessarily in this order), where $k+l=y$.
However, the $\gcd$ of the two factors on the left side is $2$, because:
Given that the product of these two factors is $5\cdot 2^y$, there are two possible cases: either one of them is equal to $2$ and the other is equal to $5\cdot 2^{y-1}$, or one of them is equal to $10$ and the other is equal to $2^{y-1}$.
In the first case, since $11^s-1 < 11^s+1$, we must have $11^s-1=2$ and $11^s+1=5\cdot 2^{y-1}$. Of course, the first equation has no solutions.
In the second case, we can argue in a similar way to get that $11^s-1=10$ and $11^s+1=2^{y-1}$. It follows from the first equation that $s=1$, however, the second equation now becomes $2^{y-1}=12$, which has no solution.
Thus, the only solution of the equation is $(x,y)=(1,1)$.