Proving any integer cube is equal to a difference of two squares 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?
 A: 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.
A: 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.
A: 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.
