# Find a couple of integers such that the third power of a given natural can be written as the difference of the squares of those integers

Given a natural number $n$, find inegers $a, b$ such that $n^3=a^2-b^2$. I've tried, but I'm a bit rusty. Please Help

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Are you trying to characterize the numbers $n$ whose cubes are equal to a difference of squares? – Dimitrije Kostic Nov 24 '11 at 18:03
Nope, given the natural number I just want to find two integers that satisfy the equation above. Not interested in finding them all. – Chu Nov 24 '11 at 18:10
What natural number $n$ are you talking about? – Dimitrije Kostic Nov 24 '11 at 18:17
Hint: $a^2-b^2=(a-b)(a+b)$. How can you make this product exactly $n^3$? – N. S. Nov 24 '11 at 18:25
@Dimitrije Kostic It is possible to do it for ALL numbers. – N. S. Nov 24 '11 at 18:26

Note that $$n^3=\left(\frac{n^2+n}{2}\right)^2-\left(\frac{n^2-n}{2}\right)^2.$$

Comment: The magic identity that solved the problem in fact did not (for me) come by magic. Given a number $K$, we want to find numbers $a$ and $b$ such that $a^2-b^2=K$. So we want $(a+b)(a-b)=K$. This means that $a+b$ and $a-b$ are two integers whose product is $K$.

Suppose that $x$ and $y$ are any two integers whose product is $K$. If we set $a+b=x$ and $a-b=y$, then we will have $(a+b)(a-b)=K$. But will $a$ and $b$ be integers?

Solve the system $a+b=x$, $a-b=y$. Algebra gives $a=\frac{x+y}{2}$, $b=\frac{x-y}{2}$. In order to make sure that $a$ and $b$ are integers, $x+y$ (and therefore $x-y$) must be even. This means that $x$ and $y$ have to be of the same parity (both odd or both even).

Can we express $n^3$ as a product of two numbers of the same parity? If $n$ is odd, we can use $x=n^3$, $y=1$. That won't work if $n$ is even. But $x=n^2$, $y=n$ always works, because $n^2$ and $n$ have the same parity.

In general, the integer $K$ is a difference of two squares unless $K$ is even but not divisible by $4$. So $\pm 2$, $\pm 6$, $\pm 10$, $\pm 14$, and so on cannot be expressed as a difference of two squares, and everybody else can be.

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@Andres: I wanted to point something out: Recall that we have the surprising identity $$\sum_{k=1}^n k^3=\left(\sum_{k=1}^n k\right)^2 =\left(\frac{n^2+n}{2}\right)^2.$$ A short proof of follows from your above identity, along with using the fact that the series telescopes. – Eric Naslund Nov 24 '11 at 19:38
Nice observation! It gives an attractive explanation of the sum of cubes formula. Alternately, we have the combinatorial version $\binom{n+1}{2}^2-\binom{n}{2}^2=n^3$. – André Nicolas Nov 24 '11 at 20:05

If $n$ is odd, then $a=(n^3+1)/2$ and $b=(n^3-1)/2$ will work. If $n$ is even, then $a=(n^3+2)/2$ and $b=(n^3-2)/2$.

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