# Show that $m^4+4n^4$ can be expressed as the sum of four squares of natural numbers

Let $m,n$ be natural numbers and $m\not=n$. Show that $m^4+4n^4$ can be expressed as the sum of four squares of natural numbers.

That is, express $m^4+4n^4$ as $A^2+B^2+C^2+D^2$ where $A,B,C,D$ is an expression in $n$ or/both $m$.

@Tim: Yes, you are completely right. This is the most careless mistake ever!! @Robert: And yes I want the sum of four squares of expressions (not necessarily polynomials, can contain the term mn) in m and/or n with integer coefficients.

And what is the significance of assuming $m\not=n?$

Any help will be greatly appreciated. Many many thanks!

• Any positive integer can be expressed as the sum of four squares (Lagrange's four-square theorem). But apparently you want something else: $m^4 + 4 n^4$ as the sum of four squares of polynomials in $m$ and $n$ with integer coefficients? Commented Dec 22, 2013 at 8:02
• Your first question can be answered by saying that you made an error. You tried to apply the sum of cubes formula to a sum of squares!
– Tim
Commented Dec 22, 2013 at 8:04
• @Tim: You are right, thanks. I don't what's in my mind... I have edited it. Commented Dec 22, 2013 at 8:17
• $(m-m)^2+(n-n)^2+(m^2)^2+(2n^2)^2.$ Commented Dec 22, 2013 at 8:46
• Thanks John! Your answer is pretty striking, it didn't come up in my mind in the first place. But I guess it will work if we assume 0 as a natural number. Commented Dec 22, 2013 at 9:01

$$m^4+4n^4=(m^2)^2+(2n^2)^2=(m^2+2n^2)^2-(2mn)^2$$ $$=(m^2+2n^2-2mn)(m^2+2n^2+2mn)=\{(m-n)^2+n^2\}\{(m+n)^2+n^2\}$$
As $m\ne n, (m-n)^2>0$