0

I'm attempting to complete the square of $m^2 + n^2$, How would I do this? I am not understanding, as most resources refer to a polynomial with $x$ as it's variable and every term is in terms of $x$.

Edit: I'm trying to do this in a proof. I'm trying to prove if $m-n$ is even, then $m^2 - n^2$ is even. My lecturer in class said that it may be easier to complete the square of the following. All I wrote down was $2*(n^2 + mn)$. Where can I go from there?

Xetrov
  • 2,089
Howard P
  • 409
  • 4
  • 16
  • 3
    Completing the square is done to "remove" the linear term from the quadratic polynomial. Here there is no linear term. The only thing I can see is writing as $(m+n)^2-2mn$ if that is helpful – imranfat Mar 01 '17 at 23:00
  • That is helpful, thank you. How did you go about doing that? – Howard P Mar 01 '17 at 23:01
  • 1
    Writing a given sum of two squares in that manner is sometimes helpful in calculus (integration). – imranfat Mar 01 '17 at 23:02
  • 1
    It would be good to see more context, but perhaps you mean $$m^2 + n^2 = \left(m + \frac{n^2}{2m}\right)^2 - \frac{n^4}{4 m^2}$$ – Robert Israel Mar 01 '17 at 23:05
  • 1
    Quite unusual! Usually, we have something like $x^2+8x$ and have to add the term $16$ to get $x^2+8x+16=(x+4)^2$ In your example, adding the term $2mn$ or the term $-2mn$ leads to $(m\pm n)^2$ – Peter Mar 01 '17 at 23:21
  • I just provided a bit more context for more understanding – Howard P Mar 02 '17 at 16:14

2 Answers2

3

When you draw a square of $n \times n$ and a square of $m \times m$ where they share one corner point, you will see that you are missing two pieces of $m \times n$ to complete an $(m+n) \times (m+n)$ square.

Pieter21
  • 3,475
0

Found the answer independently, Completing the square was the phrase my professor used, but I just needed (m^2 + n^2) in a different form. 

  (m + n)^2 - 2mn = (m^2 + n^2)

Just expand (m + n)^2 and it becomes apparent.

Howard P
  • 409
  • 4
  • 16