PRMO 2012 question 19

 

Below is exactly the chain-of-thought I would go through if this dropped onto my scratch paper in a contest or an exam. I’m writing it slowly so you can see the cues that tell me *what* to complete the square in and *why* those particular shifts appear. ---

1. Look for a **single-variable** square you can force The moment you see a quadratic in two variables, $$ x^{2}+4y^{2}-2xy-2x-4y-8=0, $$ pick one variable (almost always the one with coefficient 1 in its squared term, because that makes the algebra cleaner). Here $x^{2}$ already has coefficient 1, so treat **$y$ as a constant** for a moment and complete the square in **$x$** exactly the way you would in one-variable algebra. Write the $x$-part together: $$ x^{2} \;-\; 2xy \;-\; 2x. $$ It’s a quadratic in $x$ whose linear coefficient is $-2(y+1)$. So $$ x^{2}-2(y+1)x \;=\; \bigl(x-(y+1)\bigr)^{2} \;-\;(y+1)^{2}. $$ > **Why $y+1$?** > Because the standard completing-the-square move says: $x^{2} - 2px = (x-p)^{2}-p^{2}$. > Here $p = y+1$. Put that back into the whole equation: $$ \bigl(x-(y+1)\bigr)^{2} \;-\;(y+1)^{2} + 4y^{2}-4y-8 \;=\;0. $$ ---

2. Collect the stuff **not** in the new square Everything except the $\bigl(x-(y+1)\bigr)^{2}$ is now a function of $y$ only: $$ -(y+1)^{2}+4y^{2}-4y-8 \;=\; -\,\bigl(y^{2}+2y+1\bigr) + 4y^{2}-4y-8. $$ Do the algebra once (no tricks needed): $$ \begin{aligned} &-y^{2}-2y-1 + 4y^{2} - 4y - 8 \\ &= 3y^{2} - 6y - 9 \\ &= 3\bigl(y^{2}-2y-3\bigr) \\ &= 3\bigl((y-1)^{2}-4\bigr). \end{aligned} $$ So the equation is $$ \bigl(x-(y+1)\bigr)^{2} + 3\bigl((y-1)^{2}-4\bigr)=0 \;\;\Longrightarrow\;\; \bigl(x-(y+1)\bigr)^{2} + 3(y-1)^{2} = 12. $$ At this point you should **pause and smile**: you’ve boiled the whole thing down to *one genuine square plus another square times 3 equals a small constant*. That’s the classic signature of a manageable Diophantine-type ellipse. ---

3. See the pattern and rename cleanly It feels messy to carry $x-(y+1)$ and $y-1$ everywhere, so give them short names: $$ a:=x-y-1,\qquad b:=y-1. $$ The key is that these substitutions are *pure translations* (no rotations, no scaling), so they preserve integrality and are easy to invert. The equation is now $$ a^{2}+3b^{2}=12, $$ and that is something you can scan by eye: $3b^{2}\le12$ forces $|b|\le2$, etc. ---

4. Why this method is reliable

1. **Pick the variable with coefficient 1** in its square term; complete the square in that variable. *If neither coefficient is 1 you can still do it, you just get fractions you’ll have to clear later.*

2. **Translate, don’t rotate.** Completing the square (adding/subtracting $(\text{linear}/2)^{2}$) is a *translation* of coordinates, not a rotation, so it keeps the cross-term pattern simple.

3. **After the first square, gather what’s left.** Nine times out of ten the leftover is a single-variable quadratic that you can itself rewrite as a square ± constant.

4. **Stop when you hit “sum of (scaled) squares = constant”.** That shape almost always means a finite search (ellipse) or no solutions (negative constant) or an infinite lattice (hyperbola), and you can decide which in seconds. With practice you’ll recognize the end-form you’re aiming for—one square in each variable, maybe with a coefficient like 3 or 2 in front of one of them, equated to a *small* integer.

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5. Quick self-test Without looking up the answer, try to reduce $$ x^{2}+6y^{2}+4xy-10x+12y-15=0 $$ by the same steps. (Hint: complete the square in $x$ first, then in $y$.) If you can bring it to something like $a^{2}+3b^{2}=k$ or $a^{2}-3b^{2}=k$ on your own, you’ve internalised the technique. Good luck—once you’ve “clicked” with this translation-plus-square trick, these conic-section Diophantines become routine!