AlgebraDifficulty 25-808 tagged problems

Simon's Favorite Factoring Trick

Simon's Favorite Factoring Trick is the habit of spotting the $xy + ax + by$ shape and adding a constant so it factors as $(x+a)(y+b)$. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.

Practice this technique Match my level

The Key Result

$$xy + ax + by + ab = (x + a)(y + b)$$

Adding the right constant to a $xy + ax + by$ expression completes it into a clean product $(x+a)(y+b)$, turning a two-variable equation into a factoring problem.

Why It Works

1
Expand $(x + a)(y + b) = xy + bx + ay + ab$, so the grouped form $xy + bx + ay$ is just one constant ($ab$) short of factoring.
2
Given an equation like $xy + bx + ay = N$, add $ab$ to both sides to manufacture that missing term.
3
The left side now factors: $(x + a)(y + b) = N + ab$.
4
A product of two integers equaling a fixed number is solved by listing divisor pairs of $N + ab$ — the equation becomes a finite divisor hunt.

Worked Examples

Example 1

How many ordered pairs of positive integers $(x, y)$ satisfy $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{4}$?

1
Clear denominators by multiplying through by $4xy$: $4y + 4x = xy$, i.e. $xy - 4x - 4y = 0$.
2
Apply SFFT: add $16$ to both sides so the left factors — $xy - 4x - 4y + 16 = 16$, giving $(x - 4)(y - 4) = 16$.
3
Each ordered factorization of $16$ as a product of two positive integers gives a solution; $16 = 2^4$ has $\tau(16) = 5$ positive divisors, so there are $5$ ordered pairs $(x-4, y-4) \in \{(1,16),(2,8),(4,4),(8,2),(16,1)\}$.
4
Each yields $x, y > 4 > 0$, so all $5$ are valid (e.g. $(x,y) = (5, 20)$, $(6, 12)$, $(8, 8)$, $(12, 6)$, $(20, 5)$).

Answer:

$5$ ordered pairs.

Warm-up

Find all positive integers $(x, y)$ with $xy = x + y$.

1
Move everything to one side: $xy - x - y = 0$.
2
Add $1$ to both sides to complete the factorization: $xy - x - y + 1 = 1$, so $(x - 1)(y - 1) = 1$.
3
The only positive-integer factorization of $1$ is $1 \cdot 1$, forcing $x - 1 = 1$ and $y - 1 = 1$, i.e. $x = y = 2$.

Answer:

$(x, y) = (2, 2)$.

Contest level

How many ordered pairs of positive integers $(x, y)$ satisfy $xy - 3x - 2y = 4$?

1
To factor $xy - 3x - 2y$, note $(x - 2)(y - 3) = xy - 3x - 2y + 6$, so $xy - 3x - 2y = (x - 2)(y - 3) - 6$.
2
The equation becomes $(x - 2)(y - 3) - 6 = 4$, i.e. $(x - 2)(y - 3) = 10$.
3
Positive factor pairs of $10$ give $x > 2, y > 3$: $(x-2, y-3) \in \{(1,10),(2,5),(5,2),(10,1)\}$. Negative factor pairs force $x \le 0$ or $y \le 0$, so they are rejected.

Answer:

$4$ ordered pairs.

Olympiad / Challenge

Find the number of ordered pairs of positive integers $(x, y)$ with $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{6}$.

1
Clear denominators: $6y + 6x = xy$, i.e. $xy - 6x - 6y = 0$.
2
Apply SFFT: add $36$ to both sides — $xy - 6x - 6y + 36 = 36$, so $(x - 6)(y - 6) = 36$.
3
Each positive divisor $d$ of $36$ gives $x - 6 = d$, $y - 6 = \frac{36}{d}$ with $x, y > 6$. Since $36 = 2^2 3^2$ has $\tau(36) = (2+1)(2+1) = 9$ divisors, there are $9$ ordered solutions.

Answer:

$9$ ordered pairs.

Going Deeper

Generalization: SFFT is the additive cousin of completing the square — you add the unique constant that turns $xy + (\text{linear})$ into a product. The count of solutions to $\frac{1}{x} + \frac{1}{y} = \frac{1}{n}$ is exactly $\tau(n^2)$, the number of divisors of $n^2$.
Where it appears: AIME Diophantine counting ('how many positive integer pairs...'), unit-fraction (Egyptian-fraction) equations, and any problem where a two-variable integer equation must be reduced to a finite divisor search.
Pitfall: the constant you add is $ab$ where the linear terms are $bx + ay$ — match coefficients carefully (in $xy - 6x - 6y$ both are $-6$, so add $36$, not $6$). And once you list factor pairs, re-check the SIGN/positivity constraints: factor pairs that violate $x, y > 0$ (or whatever the domain demands) must be discarded.

Spot the Signal

Use it when a two-variable equation has an $xy$ term plus linear terms and you want integer solutions.
You can describe the hard part as spotting the $xy + ax + by$ shape and adding a constant so it factors as $(x+a)(y+b)$, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by moving everything to one side and adding the constant that completes the product.
Name the relation that makes Simon's Favorite Factoring Trick legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Forgetting to add the same constant to both sides, or mis-reading which constant completes the factorization.
Applying Simon's Favorite Factoring Trick because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Add the missing corner so the rectangle factors; then read off the divisors.

Try it on:

Find the integer solutions of a contest equation whose main obstacle is factoring an $xy + ax + by$ expression.

Back to techniques

AlgebraDifficulty 25-808 tagged problems

Simon's Favorite Factoring Trick

Practice this technique Match my level

The Key Result

$$xy + ax + by + ab = (x + a)(y + b)$$

Adding the right constant to a $xy + ax + by$ expression completes it into a clean product $(x+a)(y+b)$, turning a two-variable equation into a factoring problem.

Why It Works

1
Expand $(x + a)(y + b) = xy + bx + ay + ab$, so the grouped form $xy + bx + ay$ is just one constant ($ab$) short of factoring.
2
Given an equation like $xy + bx + ay = N$, add $ab$ to both sides to manufacture that missing term.
3
The left side now factors: $(x + a)(y + b) = N + ab$.
4
A product of two integers equaling a fixed number is solved by listing divisor pairs of $N + ab$ — the equation becomes a finite divisor hunt.

Worked Examples

Example 1

How many ordered pairs of positive integers $(x, y)$ satisfy $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{4}$?

1
Clear denominators by multiplying through by $4xy$: $4y + 4x = xy$, i.e. $xy - 4x - 4y = 0$.
2
Apply SFFT: add $16$ to both sides so the left factors — $xy - 4x - 4y + 16 = 16$, giving $(x - 4)(y - 4) = 16$.
3
Each ordered factorization of $16$ as a product of two positive integers gives a solution; $16 = 2^4$ has $\tau(16) = 5$ positive divisors, so there are $5$ ordered pairs $(x-4, y-4) \in \{(1,16),(2,8),(4,4),(8,2),(16,1)\}$.
4
Each yields $x, y > 4 > 0$, so all $5$ are valid (e.g. $(x,y) = (5, 20)$, $(6, 12)$, $(8, 8)$, $(12, 6)$, $(20, 5)$).

Answer:

$5$ ordered pairs.

Warm-up

Find all positive integers $(x, y)$ with $xy = x + y$.

1
Move everything to one side: $xy - x - y = 0$.
2
Add $1$ to both sides to complete the factorization: $xy - x - y + 1 = 1$, so $(x - 1)(y - 1) = 1$.
3
The only positive-integer factorization of $1$ is $1 \cdot 1$, forcing $x - 1 = 1$ and $y - 1 = 1$, i.e. $x = y = 2$.

Answer:

$(x, y) = (2, 2)$.

Contest level

How many ordered pairs of positive integers $(x, y)$ satisfy $xy - 3x - 2y = 4$?

1
To factor $xy - 3x - 2y$, note $(x - 2)(y - 3) = xy - 3x - 2y + 6$, so $xy - 3x - 2y = (x - 2)(y - 3) - 6$.
2
The equation becomes $(x - 2)(y - 3) - 6 = 4$, i.e. $(x - 2)(y - 3) = 10$.
3
Positive factor pairs of $10$ give $x > 2, y > 3$: $(x-2, y-3) \in \{(1,10),(2,5),(5,2),(10,1)\}$. Negative factor pairs force $x \le 0$ or $y \le 0$, so they are rejected.

Answer:

$4$ ordered pairs.

Olympiad / Challenge

Find the number of ordered pairs of positive integers $(x, y)$ with $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{6}$.

1
Clear denominators: $6y + 6x = xy$, i.e. $xy - 6x - 6y = 0$.
2
Apply SFFT: add $36$ to both sides — $xy - 6x - 6y + 36 = 36$, so $(x - 6)(y - 6) = 36$.
3
Each positive divisor $d$ of $36$ gives $x - 6 = d$, $y - 6 = \frac{36}{d}$ with $x, y > 6$. Since $36 = 2^2 3^2$ has $\tau(36) = (2+1)(2+1) = 9$ divisors, there are $9$ ordered solutions.

Answer:

$9$ ordered pairs.

Going Deeper

Generalization: SFFT is the additive cousin of completing the square — you add the unique constant that turns $xy + (\text{linear})$ into a product. The count of solutions to $\frac{1}{x} + \frac{1}{y} = \frac{1}{n}$ is exactly $\tau(n^2)$, the number of divisors of $n^2$.
Where it appears: AIME Diophantine counting ('how many positive integer pairs...'), unit-fraction (Egyptian-fraction) equations, and any problem where a two-variable integer equation must be reduced to a finite divisor search.
Pitfall: the constant you add is $ab$ where the linear terms are $bx + ay$ — match coefficients carefully (in $xy - 6x - 6y$ both are $-6$, so add $36$, not $6$). And once you list factor pairs, re-check the SIGN/positivity constraints: factor pairs that violate $x, y > 0$ (or whatever the domain demands) must be discarded.

Spot the Signal

Use it when a two-variable equation has an $xy$ term plus linear terms and you want integer solutions.
You can describe the hard part as spotting the $xy + ax + by$ shape and adding a constant so it factors as $(x+a)(y+b)$, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by moving everything to one side and adding the constant that completes the product.
Name the relation that makes Simon's Favorite Factoring Trick legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Forgetting to add the same constant to both sides, or mis-reading which constant completes the factorization.
Applying Simon's Favorite Factoring Trick because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Add the missing corner so the rectangle factors; then read off the divisors.

Try it on:

Find the integer solutions of a contest equation whose main obstacle is factoring an $xy + ax + by$ expression.