It is from the Romanian magazine "Gazeta Matematica"
Proposed for 7th grade on page 224. The statement is (thanks to Miss Google for the translation):
"Let the polynomial be given P(x,y)=by(x^2+a^2)-ax(y^2+b^2)+xy. Show that:
P(b,a)=P(-\frac{a}{4},-4b)=P(\frac{b}{a^2-b^2},\frac{a}{a^2-b^2})=P(-\frac{5ab+1}{5b},\frac{4-5ab}{5a})."
Solution CiP
It's easy to calculate P(b,a)=b \cdot a(a^2+b^2)-a \cdot b(a^2+b^2)+b \cdot a=a \cdot b.
To avoid laborious calculations, we will write the polynomial in a different way.
P(x,y)=\underline{bx^2y}+\underline{\underline{a^2by}}\; \underline{-axy^2}\;\underline{\underline{-ab^2x}}+xy,
now we group the identically-underlined terms,
P(x,y)=xy(bx-ay)+ab(ay-bx)+xy,
P(x,y)=(xy-ab)\cdot (bx-ay)+xy. \tag{1}
For the couple (x,y)=(-\frac{a}{4},-4b) we see that xy=ab so the first paranthesis in the formula (1) is equal to zero, finally obtaining the result P(x,y)=ab.
For the couple (x,y)=(\frac{b}{a^2-b^2}, \frac{a}{a^2-b^2}) we see that
bx-ay=\frac{b^2}{a^2-b^2}-\frac{a^2}{a^2-b^2}=-1.
So for these two values of x and y we have from (1) that
P(x,y)=(xy-ab)(-1)+xy=ab.
The same goes for the couple (x,y)=(-\frac{5ab+1}{5b},\frac{4-5ab}{5a}); here again
bx-ay=-\frac{5ab+1}{5}-\frac{4-5ab}{5}=\frac{-5ab-1-4+5ab}{5}=-1.
All values we calculated are equal to ab.
\blacksquare
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