The first issue I have is that I am not sure why this is called a 'compound quadratic problem', but anyway to proceed:

Suppose that $x-y=14$ and $$(x+y)(x^2+y^2)(x^4+y^4)=a(x^b-y^b)$$ where $a$ and $b$ are constants. Find $a$ and $b$.

I am stuck as to how to approach this problem. I have tried factoring (which becomes imaginary as it is the sum of squares, at which point I have no idea what to do), and expanding both sides of the equation. I had an idea that using the difference of two squares on the right hand side might get me somewhere but there is still always a term with a negative coefficient (to demonstrate what I mean): $$(x^b-y^b)=(x^\frac{b}{2}+y^\frac{b}{2})(x^\frac{b}{2}-y^\frac{b}{2})=(x^\frac{b}{2}+y^\frac{b}{2})(x^\frac{b}{4}+y^\frac{b}{4})(x^\frac{b}{4}-y^\frac{b}{4})$$

Any help appreciated.

  • 6
    $\begingroup$ Multiply it with $x-y = 14$. $\endgroup$ – Daniel Fischer Oct 10 '13 at 9:42
  • $\begingroup$ $(x^\frac{b}{2}+y^\frac{b}{2})(x^\frac{b}{4}+y^\frac{b}{4})(x^\frac{b}{8} + y^\frac{b}{8} )(x^\frac{b}{8}-y^\frac{b}{8})$ might be worth thinking about $\endgroup$ – Henry Oct 10 '13 at 10:06
  • $\begingroup$ Excellent! Somehow when I ask for help it makes the problem seem too easy.. Thanks for all help $\endgroup$ – xcvd Oct 10 '13 at 10:11

You can use this also to understand:
$$x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$$

Daniel Fischer's method is brilliant of course.

  • 1
    $\begingroup$ Thanks. Each of the solutions given led me to the correct answer, and your formula is indeed useful to understand $\endgroup$ – xcvd Oct 10 '13 at 10:15

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