When is $63\cdot2^x\pm1$ a perfect power?

Question

When is $63\cdot2^x\pm1$ a perfect power?

I noticed that:

$63\cdot2^0+1=2^6$

$63\cdot2^1-1=5^3$

$63\cdot2^8+1=127^2$

Other than $0,1,8$, are there any integer values of $x$ such that $63\cdot2^x\pm1$ is a perfect power?

Note that $63\cdot2^x-1=3$ mod $4$, so $63\cdot2^x-1$ is never a perfect square.

Answer 1

Aphelli has already provided a nice answer.

I'm going to write slightly different proofs for the following two claims :

Claim 1 : $63\cdot 2^x-1$ is a perfect power if and only if $x=1$.

Claim 2 : $63\cdot 2^x+1$ is a perfect power if and only if $x=0,8$.

Then, using the idea of the proofs of the two claims above, I'm going to write proofs to the following generalization :

Claim 3 : If $a\cdot 2^x-1\ \text{($a\ge 3$ is odd)}$ is a perfect power, then $x\lt \log_2(a+2)$.

Claim 4 : If $a\cdot 2^x+1\ \text{($a\ge 3$ is odd)}$ is a perfect power, then $x\le 2+\log_2(a+1)$.

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Claim 1 : $63\cdot 2^x-1$ is a perfect power if and only if $x=1$.

Proof :

If $x=0$, it is not a perfect power. If $x=1$, it is a perfect power.

Suppose that there are positive integers $p\ge 2,q\ge 2, x\ge 2$ such that $63\cdot 2^x-1=p^q$. Then, $p$ is odd.

• If $q$ is even, then we have $3\equiv (\underbrace{p^{q/2}}_{\text{odd}})^2\pmod 4$ which is impossible.

• If $q$ is odd, then $63\cdot 2^x=(p+1)(\underbrace{p^{q-1}-p^{q-2}+\cdots -p+1}_{\text{odd}})$, so $2^x\mid p+1$ implies $p\ge 2^x-1$. So, $63\cdot 2^x=p^q+1\ge (2^x-1)^2+1\gt 2^{2x}-2^{x+1}$ implies $63\gt 2^x-2$, so we get $2\le x\le 6$ none of which is sufficient.$\ \blacksquare$

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Claim 2 : $63\cdot 2^x+1$ is a perfect power if and only if $x=0,8$.

Proof :

If $x=0$, it is a perfect power. If $x=1$, it is not a perfect power.

Suppose that there are positive integers $p\ge 2,q\ge 2, x\ge 2$ such that $63\cdot 2^x+1=p^q$. Then, $p$ is odd.

• If $q$ is even, then $63\cdot 2^x=(\underbrace{p^{q/2}-1}_{\text{even}})(\underbrace{p^{q/2}+1}_{\text{even}})$. Since $\gcd(p^{q/2}-1,p^{q/2}+1)=2$, there are positive integers $d_1,d_2$ such that $\{p^{q/2}-1,p^{q/2}+1\}=\{2d_1,2^{x-1}d_2\}$ and $d_1d_2=63$. So, we have $p^{q/2}-1=\min(2d_1,2^{x-1}d_2)\le 2\cdot 63$. So, $63\cdot 2^x=(p^{q/2})^2-1\le (2\cdot 63+1)^2-1=4\cdot 63^2+4\cdot 63$ implies $2^x\le 4\cdot 63+4=2^8$, so we get $2\le x\le 8$ where $x=8$ is sufficient, and none of $x$ satisfying $2\le x\le 7$ is sufficient.

• If $q$ is odd, then $63\cdot 2^x=(p-1)(\underbrace{p^{q-1}+p^{q-2}+\cdots +p+1}_{\text{odd}})$, so $2^x\mid p-1$ which implies $p\ge 2^x+1$. So, $63\cdot 2^x=p^q-1\ge (2^x+1)^2-1=2^{2x}+2^{x+1}$ which implies $63\ge 2^x+2$, so we get $2\le x\le 5$ none of which is sufficient.$\ \blacksquare$

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Claim 3 : If $a\cdot 2^x-1\ \text{($a\ge 3$ is odd)}$ is a perfect power, then $x\lt \log_2(a+2)$.

Proof :

Suppose that there are positive integers $p\ge 2,q\ge 2, x\ge 1$ such that $a\cdot 2^x-1=p^q$. Then, $p$ is odd.

• If $q$ is even and $x\ge 2$, then we have $3\equiv (\underbrace{p^{q/2}}_{\text{odd}})^2\pmod 4$ which is impossible. So, if $q$ is even, then $x\le 1$.

• If $q$ is odd, then $a\cdot 2^x=(p+1)(\underbrace{p^{q-1}-p^{q-2}+\cdots -p+1}_{\text{odd}})$, so $2^x\mid p+1$ implies $p\ge 2^x-1$. So, $a\cdot 2^x=p^q+1\ge (2^x-1)^2+1\gt 2^{2x}-2^{x+1}$ implies $a\gt 2^x-2$, so we get $x\lt \log_2(a+2)$ where $\log_2(a+2)$ is larger than $1$.$\ \blacksquare$

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Claim 4 : If $a\cdot 2^x+1\ \text{($a\ge 3$ is odd)}$ is a perfect power, then $x\le 2+\log_2(a+1)$.

Proof :

Suppose that there are positive integers $p\ge 2,q\ge 2, x\ge 2$ such that $a\cdot 2^x+1=p^q$. Then, $p$ is odd.

• If $q$ is even, then $a\cdot 2^x=(\underbrace{p^{q/2}-1}_{\text{even}})(\underbrace{p^{q/2}+1}_{\text{even}})$. Since $\gcd(p^{q/2}-1,p^{q/2}+1)=2$, there are positive integers $d_1,d_2$ such that $\{p^{q/2}-1,p^{q/2}+1\}=\{2d_1,2^{x-1}d_2\}$ and $d_1d_2=a$. So, we have $p^{q/2}-1=\min(2d_1,2^{x-1}d_2)\le 2a$. So, $a\cdot 2^x=(p^{q/2})^2-1\le (2a+1)^2-1=4a^2+4a$ implies $2^x\le 4a+4$, so we get $x\le 2+\log_2(a+1)$.

• If $q$ is odd, then $a\cdot 2^x=(p-1)(\underbrace{p^{q-1}+p^{q-2}+\cdots +p+1}_{\text{odd}})$, so $2^x\mid p-1$ which implies $p\ge 2^x+1$. So, $a\cdot 2^x=p^q-1\ge (2^x+1)^2-1=2^{2x}+2^{x+1}$ which implies $a\ge 2^x+2$, so we get $x\le \log_2(a-2)$ where $\log_2(a-2)$ is smaller than $2+\log_2(a+1)$.$\ \blacksquare$

Answer 2

Suppose that $63 \cdot 2^n + 1=a^b$, with $b \geq 2$, $a$ not being a power and $n \geq 8$. Then $a$ is odd.

Assume that $b$ is odd. Then $n=v_2(a^b-1)=v_2(a-1)$, so $a-1 \geq 2^n$ and $63\cdot 2^n=a^b-1 \geq (a-1)^b \geq 2^{nb}$, so that $256^{b-1} \leq 2^{n(b-1)} \leq 63$, hence $b=1$. We get a contradiction.

Assume that $63 \cdot 2^n-1=a^b$ with $b$ odd and $n \geq 6$. Then $n=v_2(a^b+1)=v_2(a+1)$, so $a \geq 2^{n}-1$. Then $2^{n+6} \geq 63 \cdot 2^n = a^b+1 \geq (2^n-1)^b + 1 \geq 2^{b(n-1)}$, so $b(n-1) < n+6 \leq 2(n-1)$, and $b=1$. We get a contradiction.

So by your remark, we must ask when $63 \cdot 2^n=a^2-1$ for some odd $a$ with $n \geq 8$.

One of $a-1$ and $a+1$ is not divisible by $4$ but divides $63 \cdot 2^n$, so $a-1 \leq 63 \cdot 2=126$. Thus $63 \cdot 2^n=a^2-1 \leq 126 \cdot 128$, hence $n \leq 8$.