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?
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.
ReplyDeletethen 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.
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).
ReplyDeleteOh. So it should. :(
ReplyDeleteThen 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