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- Feb 14, 2012

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Initially, I thought this is another boring high school mathematics problem, but when I started to work on it, I realized I was beaten by it, with equations in variables $a$, $b$, $p$ and $q$ to which I don't see a clear way to find the values for them.

Problem:

Find all real $p$, $q$, $a$ and $b$ such that we have \(\displaystyle (2x-1)^{20}-(ax+b)^{20}=(x^2+px+q)^{10}\) for all $x$.

Attempt:

After expanding both sides of the equation using wolfram, I get

\(\displaystyle (1048576-a^{20})x^{20}-(10485760+20a^{19}b)x^{19}+(49807360-190a^{18}b^2)x^{18}+\cdots-(1140a^3b^{17}+9120)x^{3}\)

\(\displaystyle

+(-190a^2b^{18}+760)x^{2}-(20ab^{19}+40)x+1-b^{20}=x^{20}+10px^{19}+(10q^9+45p^2q^8)x^{2}+10pq^9x+q^{10}\)

And I ended up getting extremely messy equations where solving for the values for $a$, $b$, $p$ and $q$ seems impossible by equating the coefficient of $x^{20}$, $x^{19}$, $x^{18}$, $x^{3}$, $x^{2}$, $x$ and the constant...

Could anyone help me with this problem?

Thanks in advance.