I checked up to $n\leq 100000$ then found no example. So I suspect that there doesn't exist $n\in\mathbb{N}$ such that $5^n-2^n$ is a perfect square.
Then I tried to prove by modular arithmetic, but there seems no $k$ such that $\{5^n-2^n \bmod k \mid n\in\mathbb{N}\}\cap\{n^2\bmod k\mid n\in\mathbb{N}\}=\emptyset$.
Thank you for your help.

  • $\begingroup$ The only $n$ that yields a perfect square is $2$ in the Pythagorean Theorem and the number $2$ is not part of any Pythagorean triple. For this equation, the only solution is $n=0$. $\endgroup$ – poetasis Jan 7 at 20:04

Checking things modulo $5$ you can see that $n$ has to be even, as $2$ is not a square (and hence so are all odd powers). So say $n = 2m$. Then we can write $$5^n - 2^n = 5^{2m} - 2^{2m} = (5^m - 2^m)(5^m + 2^m).$$

Assume there exist $n$ such that we get perfect squares here. Then there is a minimal such $n$, that (due to laziness) we will in the following simply call $n$. Any prime that divides $5^n - 2^n$ must divide it to an even power of at least two (quickly exclude the case $5^n - 2^n = 1$ for completeness sake please). But if all these primes also divide $5^m - 2^m$ to an even power, then $5^m - 2^m$ is also a perfect square, a contradiction to the assumption that $n$ is minimal.
That means that there must exist a prime divisor $p$ of $5^n - 2^n$ that divides both $5^m - 2^m$ and $5^m + 2^m$. But then $p$ also divides the difference of the two, which is $2^{m+1}$, so $p = 2$.

But $5^n - 2^n$ is always odd, so $p = 2$ is not possible. Hence, we have found a contradiction to the assumption that any such $n$ exists.

  • 1
    $\begingroup$ Nice and clean. $\endgroup$ – Wojowu Jan 27 '20 at 9:52

If $n$ is odd and greater than $1$ then $5^n-2^n\equiv 5\pmod{8}$, so it cannot be a square. If $n$ is even then $5^{2m}-2^{2m}=a^2$ leads to a primitive Pythagorean triple $(a,2^m,5^m)$. Since $\mathbb{Z}[i]$ is a UFD all the primitive Pythagorean triples involving a power of $5$ as greatest element depend on the real and imaginary part of $(2+i)^{2m}$. The real part is always odd, while the imaginary part $a_m=\text{Im }(2+i)^{2m}$ fulfills $$ a_{m+2} = 6a_{m+1}-25 a_{m} $$ such that $$ \nu_2(a_m)\leq \nu_2(m)+1. $$ It follows that, apart from the first cases, which can be checked by hand, $a_m$ cannot be equal to $2^m$ and there is no square of the form $5^n-2^n$.


One might note that $5^n-2^n$ is a base, and every new value of $n$ brings a new prime, previously not before seen. If some prime $p$ divides $5^n-2^n$ then it divides $5^{pn}-2^{pn}$ an additional time.

So $5-2=3$ which is not square. So we replace $n$ with $3n$, to get $5^3-2^3 = 3^2 13$. But the 13 is not square, so we have to replace $n$ with $13n$. Here we now introduce two very large coprime numbers, of which the larger is a multiple of 13.

It runs away as a power of exponentials.

  • 2
    $\begingroup$ How do you show in the first line that $(5^n-2^n)$ has a prime divisor that doesn't divide $(5^i-2^i)$ for $i<n$? $\endgroup$ – Jam Jan 27 '20 at 10:41
  • $\begingroup$ Because it's an algebraic base 5/2. See numberbase.fandom.com/wiki/Algebraic_Bases where i start writing this up. There is a new factor that does not divide any lesser value, in the size of $5^{\phi(n)}$, the euler totient power. $\endgroup$ – wendy.krieger Jan 27 '20 at 12:17
  • $\begingroup$ I'm not the downvoter but I really don't see how this works. I agree that it works for $(5^n-2^n)$ up to at least $n=36$ (link) (beyond which, the numbers become too large to quickly factor) but the new factors are not all 'in the size of $5^{\phi(n)}$' or even the same order of magnitude. For $n=14$, we have a new factor of $1597$ but $5^{\phi(14)}=15625$. Or is this an asymptotic relation? Do you have any links to published proofs about this or is it all original work? $\endgroup$ – Jam Jan 27 '20 at 12:37
  • $\begingroup$ The factor that appears at $n=14$ is 11179, which is 7*1597. The values of $5^n-2^n$ is cyclic over a period of 14, reletive to mod 49, but a period of 2 relative to mod 7. If you take the fractions 0/n to (n-1)/n, these will reduce according to the common factors of d/n. Fractions that don't reduce, like 1/m, will only occur when $m\mid n$. The new factors are then the product of $5 - 2 e^{2\pi\mathrm i * m/n}$ over m coprime and smaller than n. $\endgroup$ – wendy.krieger Jan 27 '20 at 13:22
  • $\begingroup$ I see. Thank you for the clarification. Maybe you could add that to your answer and the numberbase.fandom page because I don't think it's obvious. $\endgroup$ – Jam Jan 27 '20 at 13:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.