Problem E:2718, not from AMERICAN MATHEMATICAL MONTHLY

           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 $ab$.

$\blacksquare$