Can $n$ be expressed as $a^b-c^d$, where $a,b,c,d,$ and $n$ are natural numbers, not necessary distinct?

Question

Can $n$ be expressed as $a^b-c^d$, where $a,b,c,d,$ and $n$ are natural numbers, not necessary distinct, and $b$ and $d$ can not be both equal to $1$?

For example, when $n=1319$, then $2^{11}-3^6=1319$.

What if $n=2019$?

What if $n=100!$? (! represents the factorial)

For $n=100!$, I do not need to find $a,b,c,$ and $d$. I just want to know if it is possible to be of the given form or no.

@projectilemotion That is trivial. Trivial solutions are not to be considered here. Thanks for that, I will edit my post. — Hussain-Alqatari, Jul 20 '19 at 09:52
My earlier (deleted) comment: What about the trivial case $b=d=1$? — projectilemotion, Jul 20 '19 at 09:52
$2019 = 1010^2 - 1009^2$ (you can adapt this for any odd $n$). — Adam Bailey, Jul 20 '19 at 09:58
Related to Pillai's conjecture that for every fixed $n$ only finite many solutions exist. But only the case $n=1$ is completely solved. — Peter, Jul 20 '19 at 10:04
According to this answer, it is always possible in the case $b=d=2$ for $n$ odd and $n$ divisible by $4$. — projectilemotion, Jul 20 '19 at 10:07
@projectilemotion, so useful ,,, thanks. So for $100!$ it is possible too since it is divisible by $4$. — Hussain-Alqatari, Jul 20 '19 at 10:09
Yes, indeed, we have that: $$a=\frac{100!}{4}+1,\quad c=\frac{100!}{4}-1$$ — projectilemotion, Jul 20 '19 at 10:20
If we exclude trivial case ($b,d>1$) it seems quite challenging to find $n=6$ already... — zwim, Jul 20 '19 at 10:26
$3^3-5^2$ works for n=2. $13^3-3^7$ works for $n=10$. Haven't figured out 6 yet. — Ricky Tensor, Jul 20 '19 at 11:27
$18=3^3-3^2$, $22=7^2-3^3$. $26=3^3-1^d$ for any $d$. $n=30$ is fun, $83^2-19^3=30$. $n=14$ seems like another tricky one, along with 6. — Ricky Tensor, Jul 20 '19 at 11:35
for a difference of numbers to give a third number, the numbers must be same remainder mod all of it's factors. — , Jul 20 '19 at 15:31
@Hussain-Alqatari Based on what you've stated, if I understand correctly, you can have $a = n + 1$, $b = 1$, $c = 1$ and $d = 2$. I think you may to add that $c \neq 1$ to avoid this degenerate solution. Also, as stated in Oscar Lanzi's answer, you probably would want to have neither $b$ or $d$ be $1$. — John Omielan, Jul 21 '19 at 02:04

score 2 · Answer 1 · answered Jul 20 '19 at 10:40

2

You can write any multiple of $4$ by using the identity $$(x+1)^2-(x-1)^2=4x$$Since, $100!$ is a multiple of $4$, you can write $$100!=\left(\frac{100!}{4}+1\right)^2-\left(\frac{100!}{4}-1\right)^2$$

answered Jul 20 '19 at 10:40

Anand

3,588

Oscar Lanzi · Answer 2 · 2019-07-22T20:37:39.027

1

Well, if $b$ and $d$ cannot both be equal to $1$ them $2019$ may be rendered as $2025-6=45^2-6^1$. If it is meant that neither $b$ nor $d$ can equal $1$ then you can use a difference of squares, the choice with the smallest numbers for $2019$ being $338^2-335^2$.

All odd numbers and all multiples of $4$ may be rendered as differences of squares.

edited Jul 22 '19 at 20:37

answered Jul 20 '19 at 10:10

Oscar Lanzi

39,403

Can $n$ be expressed as $a^b-c^d$, where $a,b,c,d,$ and $n$ are natural numbers, not necessary distinct?

2 Answers2