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Let $F_n$ be the $n$ th Fermat number, $F_n:=2^{2^n}+1$.

I have been working with a similar question that reads: "Prove that no Fermat number is a perfect square." Where I found an answer that reads: "$F_0 $ and $F_1$ ($3$ and $5$ respectively) are clearly not perfect squares.

For $F_n$ where $n \ge 2$, $F_n \equiv 7 \pmod {10}$. However, only numbers that are congruent to $0, 1, 4, 5, 6$, or $9 \pmod {10}$ can be perfect squares."

I do not know if this is correct and if so how to use the same method to prove the same for a $3$ rd power integer.

MathStackExchange
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joefro
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    The argument for the perfect square is valid. Ending digit $7$ rules out a perfect square. For the cubes this cannot be used , since ending digit $7$ is possible for a cube. What you can use is the (now proven) Catalan conjecture that the only consecutive perfect powers are $8$ and $9$. Since a Fermat number is $1$ more than a power of $2$, no Fermat number can be a perfect power. – Peter Aug 13 '21 at 13:07
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    The same way, you can show that a Mersenne number (number of the form $2^n-1$ with $n\ge 2$) cannot be a perfect power, the perfect square can be ruled out modulo $4$. – Peter Aug 13 '21 at 13:14
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    There is a conjecture that the Mersenne numbers and the Fermat numbers are even all squarefree, but this is not proven and might even be wrong. This is related to the Wieferich primes. – Peter Aug 13 '21 at 13:17
  • A perhaps simpler proof that a Fermat number cannot be a perfect square is by noting that, for $n \geq 1$, $2^{2^n} + 1 = (2^{2^{n-1}})^2 + 1$, which, letting $k = 2^{2^{n-1}}$, is a number clearly strictly bounded between $k^2$ and $(k+1)^2$, i.e, $k^2 < k^2 + 1 < (k+1)^2$, which implies that it is impossible for $k^2 + 1$ to be a perfect square. – egglog Aug 13 '21 at 13:36

2 Answers2

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To prove that no Fermat number can be represented as the cube of an integer, start by assuming that it can, and let $2^{2^n} + 1= k^3$, then $2^{2^n}$ can be factored into $(k-1)(k^2+k+1)$ for an integer k.

This means that $k-1 = 2^a$ and $k^2+k+1 = 2^b$ with $a+b = 2^n$

Note that $k^2 + k + 1 \text{ mod 2} = 1$ for integer values of $k$ (why?)

The solution should follow simply by noting that $2^b$ must equal $1$ by the above conclusion, which shows that no values of $k$ satisfy the equality and thus an contradiction arises.

This can be expanded to prove that no Fermat number can be represented as an odd power.

$k^{2m-1} -1 = (k-1)(\sum_{i=0}^{2m-2} k^i)$ and $\sum_{i=0}^{2m-2} k^i \text{ mod 2} = 1$

and so by a similiar argument to the one above, the proposition can be proven.

The quadratic residue proof suffices to prove no Fermat number can be represented as an even power by noting that $k^{2m} = (k^m)^2$.

Therefore, more generally, no Fermat number $2^{2^{N}} + 1$ can be represented as $k^n$ for an integer $k$ with $n \geq 2$

egglog
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Just to clarify, Fermat number is a number of the type $2^{2^n}+1$.

About the part that no Fermat number ($F_n$) is a perfect square, I think the reasoning provided is correct.

$F_n\equiv 7 \pmod {10}$ is true because of the following:

$F_n=2^{2^n}+1=(2^2)^{2^{n-1}}+1=(5-1)^{2^{n-1}}+1\equiv 2\pmod 5$ (Using binomial theorem).

Since $F_n$ is odd, $F_n\equiv 2 \pmod {5}\implies F_n \equiv 7 \pmod{10}$.

And since no square has $7$ as its last digit, $F_n$ can never be a perfect square.

However, cubes do attain every possible last digit so this same logic won't be able to show that $F_n$ is never a perfect cube.

Aditya
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    Correct, a possibility to show it for a cube is using the now proven Catalan conjecture, but maybe for a cube, there is an easier way. – Peter Aug 13 '21 at 13:10