Algebraic Manipulation

What you need to know

• Squares. $a^2-b^2=(a-b)(a+b)$ and $(a\pm b)^2=a^2\pm2ab+b^2$ (read each $\pm$ as $+$ throughout, or as $-$ throughout). Example: $2024^2-2023^2=(2024-2023)(2024+2023)=4047$.
• Cubes. $a^3+b^3=(a+b)(a^2-ab+b^2)$, $a^3-b^3=(a-b)(a^2+ab+b^2)$, and $(a+b)^3=a^3+b^3+3ab(a+b)$. Example: if $a+b=5$ and $ab=6$, then $a^3+b^3=5^3-3\cdot6\cdot5=35$. (Check: $a$ and $b$ are $2$ and $3$, and $8+27=35$.)
• Simon's Favorite Factoring Trick. $xy+ax+by+ab=(x+b)(y+a)$. So for $xy+ax+by=c$, add $ab$ to both sides to get $(x+b)(y+a)=c+ab$, then list the factor pairs of $c+ab$. Example: $xy+x+2y=10$ becomes $(x+2)(y+1)=12$.
• Three variables. $a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)$, and $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$. Example: if $a+b+c=6$ and $ab+bc+ca=11$, then $a^2+b^2+c^2=36-22=14$ (check with $1$, $2$, $3$).
• Two rarer factorizations. The Sophie Germain Identity $a^4+4b^4=(a^2+2ab+2b^2)(a^2-2ab+2b^2)$, and $x^4+x^2+1=(x^2+x+1)(x^2-x+1)$. Example: with $a=4$, $b=1$: $4^4+4=260=26\cdot10$.
• A number plus its reciprocal. Let $t=x+\frac1x$. Then $x^2+\frac1{x^2}=t^2-2$ (because $t^2=x^2+2+\frac1{x^2}$) and $x^3+\frac1{x^3}=t^3-3t$. Example: if $x+\frac1x=3$, then $x^2+\frac1{x^2}=7$ and $x^3+\frac1{x^3}=27-9=18$.
• Telescoping products. Factor every term and neighbors cancel. Example: $1-\frac1{k^2}=\frac{(k-1)(k+1)}{k\cdot k}$, so $\left(1-\frac14\right)\left(1-\frac19\right)\cdots\left(1-\frac1{100}\right)=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdots\frac{9\cdot11}{10\cdot10}=\frac12\cdot\frac{11}{10}=\frac{11}{20}$.

How AMC 10 tests it

• Problems 3–10: a number expression such as $\frac{2024^2-2023^2}{\dots}$ or $\sqrt{31\cdot30\cdot29\cdot28+1}$. Brute force is possible, but an identity makes it instant.
• Problems 8–15: "find all positive integers $x,y$ with $xy-3x+2y=20$". Use Simon's Favorite Factoring Trick, then count or add up the solutions from the factor pairs.
• Problems 10–16: given $x+\frac1x=k$, find $x^n+\frac1{x^n}$, such as $x^4+\frac1{x^4}$.
• Problems 12–18: a long product like $\frac{2^3-1}{2^3+1}\cdot\frac{3^3-1}{3^3+1}\cdots$ or $\left(1+\frac12\right)\left(1+\frac13\right)\cdots$ that telescopes after factoring.
• Problems 15–22: three unknowns with $a+b+c$, $ab+bc+ca$ and $abc$ given (or findable), asking for $a^3+b^3+c^3$ or a similar sum of powers.

Standard approaches

1. Before computing, ask whether the expression is a difference of squares, a perfect square, or a sum or difference of cubes in disguise.
2. If an equation has an $xy$ term plus $x$ and $y$ terms, rearrange it for Simon's Favorite Factoring Trick. List all factor pairs of the constant, including negative ones.
3. For $x+\frac1x$, call it $t$ and climb one power at a time: $x^{n+1}+\frac1{x^{n+1}}=t\left(x^n+\frac1{x^n}\right)-\left(x^{n-1}+\frac1{x^{n-1}}\right)$. Example: with $t=3$, $x^4+\frac1{x^4}=3\cdot18-7=47$. (Given $x-\frac1x$ instead? Square it and add $2$ to get $x^2+\frac1{x^2}$.)
4. For long products or sums, factor every term completely and write out three or four terms in a row to see what cancels.
5. If a number expression looks designed, replace the big number by $n$ and simplify with algebra. Example: $n(n+1)(n+2)(n+3)+1=(n^2+3n+1)^2$, so with $n=28$, $\sqrt{28\cdot29\cdot30\cdot31+1}=28^2+3\cdot28+1=869$.

Worked example

How many ordered pairs of integers $(x,y)$ satisfy $xy+4x-3y=30$?

(A) $6$ (B) $8$ (C) $10$ (D) $12$ (E) $18$

The first two terms share an $x$, so write the equation as $x(y+4)-3y=30$. To get a second $(y+4)$ we need $-3(y+4)=-3y-12$, so subtract $12$ from both sides:
$$x(y+4)-3(y+4)=18\quad\Longrightarrow\quad (x-3)(y+4)=18 .$$
Now $x-3$ and $y+4$ are integers that multiply to $18$. Each way to write $18=u\cdot v$ with integers $u$, $v$ gives exactly one solution, $x=u+3$ and $y=v-4$. For example, $u=2$, $v=9$ gives $(5,5)$, and indeed $25+20-15=30$.

Count the pairs $(u,v)$. Since $18=2\cdot3^2$, it has $(1+1)(2+1)=6$ positive divisors: $1,2,3,6,9,18$. Each one can be $u$, with $v=\frac{18}u$, so there are $6$ ordered pairs with both factors positive. Making both factors negative gives $6$ more, such as $(-1,-18)$. That is $12$ pairs in all.

The answer is $\boxed{\textbf{(D)}\ 12}$.

Pitfalls

• Forgetting the negative factor pairs in Simon's Favorite Factoring Trick, which cuts the count in half. Or adding the extra constant to only one side.
• Sign mistakes between $a^3-b^3$ and $a^3+b^3$, or in $(a-b)^2=a^2-2ab+b^2$. Example: $a^3+b^3$ has $-ab$ in its second factor, not $+ab$.
• Multiplying out a designed number expression by hand instead of replacing the big number with a letter.
• In telescoping products, cancelling correctly but losing track of which terms survive at each end. Example: in $\frac21\cdot\frac32\cdot\frac43\cdots\frac{n+1}n$, only the last top and the first bottom survive, leaving $n+1$.

Traps that recur

• Trying to reduce 2^202 by finding the order of 2 modulo the divisor, or forgetting to check that 201 is smaller than the divisor. (2020 AMC 10B #22)
• Trying to solve the system directly or subtracting the equations instead of adding them; the sum is the one combination that collapses to a sum of squares. (2012 AMC 10A #24)
• Trying to solve the two equations for a, b, c individually; the system has infinitely many solutions, and the target expression is the same for all of them. (2002 AMC 10B #20)
• Using 13 as a + b + c (each edge length appears 4 times, so a + b + c = 13/4) or 11/2 as ab + bc + ca instead of 11/4. (2023 AMC 10B #17)