## Wednesday, January 17, 2007

### Ugly! Topic: Algebra. Level: AIME.

Problem: (2001 Putnam - B2) Find all pairs of real numbers $(x,y)$ satisfying the system of equations

$\displaystyle \frac{1}{x}+\frac{1}{2y} = (x^2+3y^2)(3x^2+y^2)$ and $\displaystyle \frac{1}{x}-\frac{1}{2y} = 2(y^4-x^4)$.

Solution: Add and subtract the two equations to get two new equations

$\frac{2}{x} = x^4+10x^2y^2+5y^4$ and $\frac{1}{y} = y^4+10y^2x^2+5x^4$,

which look curiously similar. Anyway, multiply through by $x$ and $y$, respectively, to find

$2 = x^5+10x^3y^2+5xy^4$ and $1 = y^5+10x^2y^3+5x^4y$.

Hmm, $1, 5, 10, \ldots$. Those look suspiciously like binomial coefficients. Add and subtract these two equations as well, resulting in

$3 = x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5 = (x+y)^5$

$1 = x^5-5x^4y+10x^3y^2-10x^2y^3+5xy^4-y^5 = (x-y)^5$.

Well, taking the fifth roots and solving the easy linear system, we have determined that the only solution is $\displaystyle (x,y) = \left(\frac{\sqrt[5]{3}+1}{2} \mbox{ }, \frac{\sqrt[5]{3}-1}{2}\right)$. QED.

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Comment: Nice how these two ugly equations reduce to two simple equations through a process of no more than three steps. Of course, working it out required a lot more than that unless you just luckily stumbled upon the answer. This is a good exercise is recognizing familiar factorizations (the binomial theorem, Sophie Germain identity, and $a^3+b^3+c^3-3abc$ come to mind), which can often result in a clean solution to a problem.

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Practice Problem: (2001 Putnam - A3) For each integer $m$, consider the polynomial

$P_m(x) = x^4-(2m+4)x^2+(m-2)^2$.

For what values of $m$ is $P_m(x)$ the product of two nonconstant polynomials with integer coefficients?

1. let p(x) have linear factor x - c; c^2 is a root of t^2 - (2m+4) t + (m-2)^2 which has solutions c^2 = m+2 \pm 2 sqrt(2m). let m = 2n^2, then c^2 = 2(n \pm 1)^2 which has only the trivial solution n = \mp 1, c = 0, m = 2.

then consider p(x) = (x^2 - ax + b)(x^2 + ax + b) = x^4 + (1 - a^2) x^2 + b^2, then b = m-2 so that 1 - a^2 = 2b + 8, or b = 2-m so that 1 - a^2 = 8 - 2b. (it's not hard to show this is the only possible quadratic factoring; this is a generalization of sophie-germaine)

either way, all odd a = 2k+1 produce a solution that gives m = whatever.

2. Eh? How did you expand (x^2-ax+b)(x^2+ax+b)? Shouldn't the x^2 term be (2b-a^2), not (1-a^2).

3. Oh. So it should. :(

Then we have 2b - a^2 = 2b + 8 (no integer solutions... or real ones, for that matter) or 2b - a^2 = 8 - 2b which has solutions for all even a = 2k. In particular,

b = 2 + k^2
m = -k^2