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I know this problem has been asked before by someone. However, my problem is a bit different. I found the following problem in Elementary number theory by Burton:

Prove that the cube of any integer can be written as the difference of two squares. Notice that $n^3 = (1^3+2^3+3^3+.......+n^3) - (1^3+2^3+3^3+.......+(n-1)^3).$

My question: can this be done using the induction method? I know that:

$n^3 = (\frac{n(n+1)}{2})^2 - (\frac{(n-1)(n)}{2})^2$

I am stuck in the induction step. Can someone help?

jvdhooft
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Nimantha
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    If you know that $n^3=\left(\frac{n(n+1)}2\right)^2-\left(\frac{n(n-1)}2\right)^2$, why do you need induction? – robjohn Apr 07 '18 at 06:25
  • I interpretted this to mean that the op was trying to prove $n^3=\left(\frac{n(n+1)}2\right)^2-\left(\frac{n(n-1)}2\right)^2$ via induction but got stuck. But @robjohn has a point. The "notice" implies that if we know $\sum k^3$ is a perfect square we'd be done. And the op's fact implies that $\sum k^3 = (\sum k)^2$. Frankly, I've never come across that before. It's easy to prove it with induction but I'm not seeing a direct proof off the top of my head. – fleablood Apr 07 '18 at 17:50
  • @fleablood: Proving the identity $\sum_{k=1}^n {k^3} = \big(\sum_{k=1}^n k\big)^2$ without induction has several proofs of this equation (including one of mine ;-). – robjohn Apr 07 '18 at 19:15
  • I don't doubt it. I've just come up with three of my own. The point is, once we know and accept it (and it's easy to learn and accept, although perhaps not jump in you face obvious) then every cube is a difference of squares is trivial. (As $n^3$ = sum to n of $k^3$ - sum to n-1 of $k^3$). – fleablood Apr 07 '18 at 20:28

3 Answers3

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Well, what happens when you try?

$(n+1)^3 = n^3 + 3n^2 + 3n + 1 =$

$(\frac {n(n+1)}2)^2 - (\frac {n(n-1)}2)^2 + 3n^2 + 3n + 1$

Meanwhile $(\frac{(n+1)(n+2)}2)^2 - (\frac {(n+1)n}2)^2 =$

$(\frac{(n+1)n}2 + \frac {2(n+1)}{2})^2 - (\frac {n(n-1)}2 + \frac {2n}{2})^2=$

$(\frac {(n+1)n}2)^2 + (n+1)^2n +(n+1)^2 - (\frac {n(n-1)}2)^2 -n^2(n-1) - n^2=$

$(\frac {(n+1)n}2)^2- (\frac {n(n-1)}2)^2 + (n+1)^2(n+1) - n^2(n)$

$(\frac {(n+1)n}2)^2- (\frac {n(n-1)}2)^2 + (n+1)^3 - n^3$.

$(\frac {n(n+1)}2)^2 - (\frac {n(n-1)}2)^2 + n^3 +3n^2 + 3n + 1- n^3=$

$(\frac {n(n+1)}2)^2 - (\frac {n(n-1)}2)^2 + +3n^2 + 3n + 1$.

So that's that.

fleablood
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  • Out of all the answers here, this is the only one that shows all the math(s) behind it. I would $+1$ if I didn't reach my daily voting limit :) – Mr Pie Apr 07 '18 at 08:12
  • Well, to be fair, the hint implies $\sum k^3$ is always a perfect square. Which if I felt that was common knowledge I'd be persuaded to pursue. However the OP wanted to do induction to show $n^3 = (\frac {n(n+1)}2)^2 - (\frac {n(n-1)}2)^2$ and the OP could continue. So I feel the best answers often are to ask "well, what would happen if you continued through?" – fleablood Apr 07 '18 at 08:37
  • @fleablood thank you very much.And I would like to know, Is trying to prove that "cube of any integer can be written as a difference of squares" using induction something useless?? The answer below, by coffeemath tells that no induction is needed.Is that true?? – Nimantha Apr 07 '18 at 13:57
  • Just because there are two ways doesn't mean one way ofproving something is "useless". Seldom is induction the only way to prove something. Coffeemath answer begins with assuming that $\sum k^3 = (\sum k)^2$ is common knowledge but doesn't prove it. If I were to assume $\sum k = \frac {n(n+1)}2$ were common knowledge then your proposition $n^3 = (\frac {n(n+1)}2)^2 - (\frac {n(n-1)}2)^2= (\sum^n k)^2 - (\sum^{n-1} k)^2$ is exactly the same proposition as coffeemaths assumtion. We still must prove that $(\sum k)^2 = \sum k^3$. Maybe we don't need induction but induction is usually easy. – fleablood Apr 07 '18 at 17:37
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I think what you're supposed to pick up here is that $\sum_1^x n^3$ is equal to $\frac{x^4 + 2n^3 + n^2}{4}$. Subtract and see what you get.

user361424
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Since the sum of the first $n$ cubes is the square of the sum of the first $n$ integers, the hint shows $n^3$ to be a difference of squares. I guess you already knew that in your post, since you gave the formula. (Of course you know the two fractional expressions are integers, since the tops are products of adjacent integers, one of which is even.) Note there is no induction to be done.

coffeemath
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