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If $x+y=m$ and $x-y=n $,

Show that $16(x^4-7x^2y^2+y^4)=(5m^2-n^2)(5n^2-m^2)$

$x^4-7x^2y^2+y^4=(x^2-y^2)^2-5x^2y^2$

$\Rightarrow [(x+y)(x-y)]^2-\dfrac{1}{5}(5xy)^2$

Then I'm stuck. The problem is the $5xy$ term has an odd coefficient, so I think no matter how I manipulate $m$ and $n$ (adding/subtracting $m^2$ and $n^2$ for example), I cannot produce an odd $xy$ term. I suppose I could just do $x=\dfrac{m+n}{2}$ and $y=\dfrac{m-n}{2}$, but that seems tedious and not the correct way to solve this problem .

Any help is appreciated. Thank you.

ronald christenkkson
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1 Answers1

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I'm not sure what made you stop as you were tantalizingly close to the solution. Note that using the expressions you derived for $x$ and $y$ gives

$$\begin{align} xy &= \left( \frac{m+n}{2} \right) \left( \frac{m-n}{2} \right) \\[2mm] &= \frac{m^2-n^2}{4} \end{align}$$

Now going back to your earlier equation (and not factoring out the $\frac{1}{5}$ from the last term) we get

$$\begin{align} x^4 - 7x^2 y^2 + y^4 &= (x+y)^2 (x-y)^2 - 5x^2 y^2 \\[2mm] &= m^2n^2 - 5 \left( \frac{m^2-n^2}{4} \right)^2 \\[2mm] &= \frac{16m^2n^2}{16} - \frac{5(m^4 - 2m^2n^2 + n^4)}{16} \\[2mm] &= \frac{-5m^4 + 26 m^2n^2 -5n^4}{16} \\[2mm] &= \frac{(5m^2-n^2)(5n^2-m^2)}{16} \end{align}$$

A.J.
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