... sorry, Diophantine !
This is Exercise S:E24.305 from a magazine dear to us. The authors are Răzvan LUPU and Roxana VASILE from CRAIOVA. The exercise is proposed for 7th grade. In translation:
"Determine the integers x and y that verify the relationship
x^2-xy+2x-3y=6."
ANSWER CiP
(x,y)\in \{(0,-2),\;(-2,-6),\;(-4,-2),\;(-6,-6)\}
Solution CiP
We write the equation successively
x^2+3x-x-3+3-xy-3y=6\;\;\Leftrightarrow\;\;x(x+3)-(x+3)+3-y(x+3)=6\;\;\Leftrightarrow
\Leftrightarrow \;\;(x+3)(x-1-y)=3
Because (x,y)\in \mathbb{Z}\times \mathbb{Z}, we have the cases indicated on the columns in the table:
\begin{array}{c||c|c|c|c}x+3&1&-1&3&-3\\ \hline x-1-y&3&-3&1&-1\\ \hline \hline x&-2&-4&0&-6 \\ \hline y&-6&-2&-2&-6 \end{array}
We got the answer.
\blacksquare
Remark CiP We guessed the decomposition by following the procedure of completing squares.
x^2-xy+2x-3y=x^2-x(y-2)-3y=x^2-2\cdot x\cdot \frac{y-2}{2}-3y=
=\left (x-\frac{y-2}{2}\right )^2-\underline{\left ( \frac{y-2}{2} \right )^2}-3y=\left (x-\frac{1}{2}y+1 \right)^2-\frac{1}{4}y^2-2y-1=
=\left(x-\frac{1}{2}y+1 \right )^2-\frac{1}{4}\cdot (y^2+8y+16)+\underline{4}-1=\left (x-\frac{1}{2}y+1 \right )^2-\frac{1}{4}(y+4)^2+3=
=\left [ \left (x-\frac{1}{2}y+1\right )-\frac{1}{2}(y+4) \right ]\cdot \left [ \left (x-\frac{1}{2}y+1 \right )+\frac{1}{2}(y+4) \right ]+3=
=(x-y-1) \cdot (x+3)+3...
<end Rem>
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