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With paper-and-pencil method I found only a first $5$ cases:

$$1=3^2-2^3$$

$$2=3^3-5^2$$

$$3=2^7-5^3$$

$$4=5^3-11^2$$

$$5=2^5-3^3$$

This looks interesting and if a natural $n$ can be represented as difference of two powers (we do not take here $a^1$ into consideration but only exponents $\geq 2$ and we do not take into consideration powers $1^m$) we can call $n$ a power-representable natural number.

It is very reasonable to expect that some numbers can be represented in more than one way but I would like to know here is it known to be true and is it true a following statement:

Every natural number is power-representable.

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    $\begingroup$ You can start by showing that every positive integer that's not of the form $4n+2$ is a difference of squares. $3 = 2^2 - 1^2$ and $5 = 3^2-2^2$ are much simpler relations that the ones you found. $\endgroup$
    – Ethan Bolker
    Commented Mar 21, 2018 at 12:07
  • $\begingroup$ @EthanBolker I would like to exclude $1$. Edit comes soon. $\endgroup$
    – Shalom
    Commented Mar 21, 2018 at 12:12
  • $\begingroup$ Excluding $1$ still gets you all the odds after $3$ and all the doubly evens after $8$. Counting the number of representations of $n$ as a difference of squares is a well known solved problem. The argument relies on the identity $a^2 - b^2 = (a-b)(a+b)$, which leads you to think about the ways you can factor $n$. $\endgroup$
    – Ethan Bolker
    Commented Mar 21, 2018 at 12:14
  • $\begingroup$ @EthanBolker I do not see that, can you write a partial(complete) answer? $\endgroup$
    – Shalom
    Commented Mar 21, 2018 at 12:17
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    $\begingroup$ Your profile shows that you are an avid new student of number theory, so exploring and generating conjectures. That's good. But your questions are based on very few examples. You can't reliably guess at properties of all the integers when you haven't even looked at all the ones less than $100$. So do keep exploring, but when you get an idea, test it more thoroughly and then try to prove it for yourself before asking here. $\endgroup$
    – Ethan Bolker
    Commented Mar 21, 2018 at 12:42

2 Answers 2

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Differences of squares are well understood.

If $$ n = a^2 - b^2 = (a-b)(a+b) $$ then $a-b$ and $a+b$ are either both even or both odd, so $n$ is either odd or a multiple of $4$.

Suppose $n$ is odd. Then each way to write $n = rs$ as a product of two (necessarily odd) factors with $r > s$ tells you $$ n = \left( \frac{r+ s}{2} \right)^2 - \left( \frac{r- s}{2} \right)^2 . $$

You can always take $r=n= 2k+1$ and $s=1$, to get the well known $$ 2k+1 = (k+1)^2 - k^2 . $$

If $n$ is prime that's the only way to write it as a difference of squares.

I leave it to you to find all the ways to write $105 = 3 \times 5 \times 7$.

Then you can work out the argument for the multiples of $4$.

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  • $\begingroup$ I've just written up the "multiples of 4" thing to flesh out what Ethan's written here. $\endgroup$
    – John Hughes
    Commented Mar 21, 2018 at 12:34
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A partial answer since as @Ethan points out, the problem is well-known and solved in general.

I'm going to write an integer of the form $4n$ as a difference of squares. A slightly more subtle argument works for $4n + 1$ and $4n + 3$.

Suppose that you look at an integer $k = 4n$, where $n$ is a positive integer. $k$ is not prime, for $k = 2(2n)$. If we write $$ a + b = 2n a - b = 2 $$ we have two equations in two unknowns; adding, we get $$ 2a = 2n + 2 $$ so $$ a = n + 1 $$ Similarly, $b = n-1$.

That gives us two integers, $a$ and $b$, with the property that $(a+b)(a-b) = 4n$. But $(a+b)(a-b) = a^2 - b^2$, so our number $k = 4n$ is a difference of squares.

As an example, look at $k = 4\cdot 5 = 20$, so $n = 5$. The formula above says to pick $a = 6$ and $b = 4$. We compute $$(a+b)(a-b) = 10 \cdot 2 = 20.$$ But this is the same as $$ a^2 - b^2 = 36 - 16 = 20 $$ so we've written $20$ as a difference of squares.

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  • $\begingroup$ I note that you marked your answer community wiki.- I just did that for mine. No one should get rep here for recording this well known material. $\endgroup$
    – Ethan Bolker
    Commented Mar 21, 2018 at 12:38
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    $\begingroup$ I think there's a case to be made for credit for writing down well-known stuff in a way comprehensible to the OP; otherwise almost no one would deserve credit for anything on MSE. :) I tend to use Community Wiki when I write an answer made up from comments by others; I do that in hopes that the single answer will be more readable, and that OP will accept it, thus closing out the question. (In this case, I was lazy and only did one case. :) ) $\endgroup$
    – John Hughes
    Commented Mar 21, 2018 at 13:30
  • $\begingroup$ But I also us C-W when something seems really basic, and just looks like a silly lapse in OP's thinking, a question like "Find a vector $v$ that lies in every subspace of some vector space $V$," i.e., for things where the answer is likely to make the OP slap their forehead and say "D'oh!". $\endgroup$
    – John Hughes
    Commented Mar 21, 2018 at 13:32

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