The Difference (and Sum) of Two Powers

We know that the difference of two squares can be factorised as follows:


So for example: \[9x^2-16=(3x-4)(3x+4)\]

You may wonder if we can extend this to cubes and other powers. In fact, we can factorise the difference of any powers as follows:

\[a^n-b^n=(a-b)(a^{n-1}+ba^{n-2}+b^2a^{n-3}+ \cdots +b^{n-3}a^2+b^{n-2}a+b^{n-1})=(a-b)\left(\sum_{r=0}^{n-1}a^{n-1-k}b^k\right)\]

You may also wonder if we can factorise the sum of two powers. This only works for odd powers, and can be done as follows:

\[a^{2n+1}+b^{2n+1}=(a+b)(a^{2n}-ba^{2n-1}+b^2a^{2n-2}- \cdots +b^{2n-2}a^2-b^{2n-1}a+b^{2n})=(a+b)\left(\sum_{r=0}^{2n} a^{2n-k}(-b)^k\right)\]

Note that this time the signs in the factorisation alternate, starting with a \(+\) in the first bracket, then a \(-\) as the first sign in the second bracket, a \(+\) as the second and so on.

Proof of \(a^n-b^n\)

We can prove the above quite easily by considering the right-hand side of the equations. For the first equation, we have:


Expanding gives: \[a\sum_{r=0}^{n-1}(a^{n-1-k}b^k) -b\sum_{r=0}^{n-1}(a^{n-1-k}b^k)\]

\[=\sum_{r=0}^{n-1}(a^{n-k}b^k) -\sum_{r=0}^{n-1}(a^{n-1-k}b^{k+1})\]

Notice that the second term is just the first term but with \(k+1\) instead of \(k\), so we can rewrite the second term as:

\[\sum_{r=0}^{n-1}(a^{n-k}b^k) -\sum_{r=1}^{n}(a^{n-k}b^{k})\]

Now notice that all of the terms accept \(a^n\) and \(b^n\) cancel, giving \(a^n-b^n\).

Proof of \(a^{2n+1}+b^{2n+1}\)

We can use a similar approach:

\[(a+b)\left(\sum_{r=0}^{2n} a^{2n-k}(-b)^k\right)\]

\[=a\sum_{r=0}^{2n} (a^{2n-k}(-b)^k)+b\sum_{r=0}^{2n} (a^{2n-k}(-b)^k)\]

\[=a\sum_{r=0}^{2n} (a^{2n-k}(-b)^k)-(-b)\sum_{r=0}^{2n} (a^{2n-k}(-b)^k)\]

\[=\sum_{r=0}^{2n} (a^{2n-k+1}(-b)^k)-\sum_{r=0}^{2n} (a^{2n-k}(-b)^{k+1})\]

\[=\sum_{r=0}^{2n} (a^{2n-k+1}(-b)^k)-\sum_{r=0}^{2n} (a^{2n-k-1+1}(-b)^{k+1})\]

\[=\sum_{r=0}^{2n} (a^{2n-k+1}(-b)^k)-\sum_{r=1}^{2n+1} (a^{2n-k+1}(-b)^{k})\]




We could then, conjecture that we can write \(a^{2n}+b^{2n}\) in the form \((a+b)(p)\), where \(p\) is an integer if both \(a\) and \(b\) are integers. However, we can easily disprove this by counterexample.  For example, take \(a=1\) and \(b=2\). Then:



\[\therefore p=\frac{5}{3}\]

But this is not an integer, so our conjecture is false.

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