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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.

user1551
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IONELA BUCIU
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    The only non-negative integer $y\le 10^5$ for which $5\cdot 2^y+1$ is a perfect power , is $4$. This gives however $3^4$. Hence , I guess the only solution is $(1/1)$ – Peter Aug 08 '23 at 16:26
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    Subtract 1 and get $11^x-1=5\cdot 2^y$. Now use the formula $11^x-1=(11-1)(11^{x-1}+\cdot+11+1)$ – Valerio Aug 08 '23 at 16:30
  • @Valerio How will this help? – IONELA BUCIU Aug 08 '23 at 16:32
  • It's the formula for the geometric sum $a^{n-1}+\cdots+a^2+a+1=\frac{a^n-1}{a-1}$ but use 11 in place of $a$ and $x$ in place of $n$ – Valerio Aug 08 '23 at 16:35
  • @Valerio Oooo...now i understand! Thank you so much for the idea! – IONELA BUCIU Aug 08 '23 at 16:37
  • If you simplify you get $10(11^{x-1}+\cdots+11+1)=5\cdot 2^y$. Divide by 5 and add $11+1=12$ and get $2(11^{x-1}+\cdots +12)=2^y$. Not that if $x>1$, then the number in parenthesis is odd, but $2^y$ has no odd factors. So the only possibility is that $x=1$ and $y=1$ – Valerio Aug 08 '23 at 16:39
  • @Valerio Why is the sum odd for $x>1$ ? Does it not contain powers of $11$ apart from the last term which is $12$ ? – Peter Aug 08 '23 at 16:50
  • @Peter Yes if $x$ were even, then the quantity in parentheses would be even. But tin this case, replacing $x$ with $2x$, we get $121^x -1 =5\cdot 2^x$ and then as before we find $24(121^{x-1}+\cdots+122)=2^y$ which would show that $2^y$ has 3 as a factor. – Valerio Aug 08 '23 at 17:32

1 Answers1

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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:

  • They are both even.
  • If an integer $d$ divides both of them, then it should also divide their difference, which equals $2$.

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 $(x,y)=(1,1)$.

George Gavrilopoulos
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