Chapter 7 — Special Factorizations & Clever Manipulations
Topics in this chapter:
7.1 Factorizations
The Four Key Factorizations
| # | Identity | Name |
|---|---|---|
| 1 | $a^2 - b^2 = (a-b)(a+b)$ | Difference of squares |
| 2 | $a^2 + b^2 = (a+b)^2 - 2ab$ | Sum of squares (rearranged) |
| 3 | $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$ | Difference of cubes |
| 4 | $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$ | Sum of cubes |
Memory aid for cubes: “Same, Opposite, Always Positive” (SOAP).
$a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)$
The first factor has the same sign; the middle term of the second factor has the opposite sign; the last term of the second factor is always positive.
Sophie Germain Identity
Derivation: $x^4 + 4y^4 = (x^2 + 2y^2)^2 - (2xy)^2$, then apply difference of squares.
Telescoping via Factorization
Example 7-1: Find $\displaystyle\sum_{k=1}^{n} \frac{1}{k(k+1)}$.
Solution: Partial fractions: $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$.
The sum telescopes:
\[\left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right) = 1 - \frac{1}{n+1} = \frac{n}{n+1}\]Example 7-2: Evaluate $1001 \cdot 1003 - 1002^2$.
Solution: Let $x = 1002$. Then:
\[(x-1)(x+1) - x^2 = x^2 - 1 - x^2 = -1\]Example 7-3: Simplify $\dfrac{1}{\sqrt{5} - \sqrt{3}}$.
Solution: Multiply by the conjugate $\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}$:
\[\frac{\sqrt{5}+\sqrt{3}}{(\sqrt{5})^2 - (\sqrt{3})^2} = \frac{\sqrt{5}+\sqrt{3}}{2}\]This technique — rationalizing the denominator — is an application of difference of squares.
Exercise 7-1: Factor completely:
i. $x^3 - 27$ ii. $8a^3 + 125b^3$ iii. $x^4 + 4$ (Hint: Sophie Germain) iv. $x^6 - 1$
Exercise 7-2:
i. Evaluate $\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \cdots + \frac{1}{99 \cdot 101}$.
ii. Show that $\frac{1}{\sqrt{k} + \sqrt{k+1}} = \sqrt{k+1} - \sqrt{k}$, and use this to evaluate $\displaystyle\sum_{k=1}^{99} \frac{1}{\sqrt{k}+\sqrt{k+1}}$.
7.2 Manipulations
Computing Higher Powers Without Solving
Key tools:
\[(x + y)^2 = x^2 + y^2 + 2xy \implies x^2 + y^2 = (x+y)^2 - 2xy\] \[(x + y)^3 = x^3 + y^3 + 3xy(x+y) \implies x^3 + y^3 = (x+y)^3 - 3xy(x+y)\]These let you find $x^2 + y^2$, $x^3 + y^3$, etc. from just $x + y$ and $xy$.
Example 7-4: If $x + y = 3$ and $xy = 1$, find $x^2 + y^2$ and $x^3 + y^3$.
Solution:
\[x^2 + y^2 = (x+y)^2 - 2xy = 9 - 2 = 7\] \[x^3 + y^3 = (x+y)^3 - 3xy(x+y) = 27 - 9 = 18\]Example 7-5: If $x + y = 5$ and $x^2 + y^2 = 13$, find $x^4 + y^4$.
Solution: First find $xy$:
\[xy = \frac{(x+y)^2 - (x^2+y^2)}{2} = \frac{25 - 13}{2} = 6\]Now find $x^4 + y^4$:
\[x^4 + y^4 = (x^2 + y^2)^2 - 2(xy)^2 = 169 - 72 = 97\]Example 7-6: If $a + b = 8$ and $ab = 12$, find $a^4 + b^4$.
Solution: Step by step:
\[a^2 + b^2 = (a+b)^2 - 2ab = 64 - 24 = 40\] \[a^4 + b^4 = (a^2 + b^2)^2 - 2(ab)^2 = 1600 - 288 = 1312\]Simon’s Favorite Factoring Trick
When you have an expression like $xy + ax + by + c$, try adding and subtracting a constant to make it factorable:
\[xy + ax + by + ab = (x + b)(y + a)\]So: $xy + ax + by + c = (x+b)(y+a) + (c - ab)$.
This is invaluable when $x$ and $y$ must be integers.
Example of SFFT: Find positive integers $x, y$ with $xy + 3x + 2y = 17$.
Add $6$ to both sides: $xy + 3x + 2y + 6 = 23$, so $(x+2)(y+3) = 23$.
Since $23$ is prime: $x + 2 = 1, y + 3 = 23$ (gives $x = -1$, invalid) or $x + 2 = 23, y + 3 = 1$ (gives $y = -2$, invalid).
So there are no positive integer solutions.
Exercise 7-3: Given $x + y = 7$ and $xy = 5$, find:
i. $x^2 + y^2$ ii. $x^3 + y^3$ iii. $x^4 + y^4$
Exercise 7-4: Given $a + b + c = 6$, $ab + bc + ca = 11$, and $abc = 6$:
i. Find $a^2 + b^2 + c^2$. ii. Find $a^3 + b^3 + c^3$.
(Hint: $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)$ and use Newton’s identity or expand $(a+b+c)^3$.)
Exercises & Problems
End-of-Chapter Problems
Problem 138. Evaluate $999^2 - 998^2 + 997^2 - 996^2 + \cdots + 3^2 - 2^2 + 1^2$.
Problem 139. Factor $a^4 - b^4$ completely.
Problem 140. If $a + b = 4$ and $ab = -3$, evaluate $a^2 + b^2$, $a^3 + b^3$, $a^4 + b^4$.
Problem 141. Simplify $\dfrac{x^3 - 8}{x^2 - 4}$.
Problem 142. Evaluate $\dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dfrac{1}{3 \cdot 4} + \cdots + \dfrac{1}{2000 \cdot 2001}$.
Problem 143. If $r$ and $s$ are roots of $x^2 - 7x + 4 = 0$, find $r^2 + s^2$ and $r^3 + s^3$. (MAΘ 1987)
Problem 144. Express $(a^2 + b^2)(c^2 + d^2)$ as the sum of two squares. (This is related to the Brahmagupta–Fibonacci identity.)
Problem 145. Simplify $\dfrac{1}{\sqrt{7}-\sqrt{5}} + \dfrac{1}{\sqrt{5}-\sqrt{3}} + \dfrac{1}{\sqrt{3}-1}$.
Problem 146. Factor $n^5 - n$ (for integer $n$) and show it is always divisible by 30.
Problem 147. Evaluate: $\displaystyle\prod_{k=2}^{n} \frac{k^3-1}{k^3+1}$.
Problem 148. If $x + \frac{1}{x} = 3$, find $x^5 + \frac{1}{x^5}$.
Problem 149. Factor $x^4 + x^2 + 1$ over the integers. (Hint: add and subtract $x^2$.)
Problem 150. If $a + b + c = 0$, simplify $\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab}$.
Problem 151. Compute $\sqrt{2 + \sqrt{2 + \sqrt{2 + \cdots}}}$. (Hint: set the expression equal to $x$.)
Problem 152. If $(x + y)^2 = 4$ and $xy = -1$, find $x^4 + y^4$.
Problem 153. For what $n$ is $1 + 2 + \cdots + n$ a perfect square?
Problem 154. Prove: $n^4 + 4^n$ is composite for all integers $n > 1$. (Hint: consider odd and even $n$ separately; use Sophie Germain.)
Problem 155. Simplify $\sqrt{6+\sqrt{20}} - \sqrt{6-\sqrt{20}}$.
Problem 156. Find all $(x,y,z)$ with $x + y + z = 6$, $xy + yz + zx = 11$, $xyz = 6$.
Problem 157. Evaluate $\dfrac{1}{1\cdot 2\cdot 3} + \dfrac{1}{2\cdot 3\cdot 4} + \cdots + \dfrac{1}{18\cdot 19\cdot 20}$.
Problem 158. If $a^2 + b^2 + c^2 = ab + bc + ca$, prove $a = b = c$. (Hint: multiply both sides by 2 and rearrange.)
Problem 159. Factor $x^3 + y^3 + z^3 - 3xyz$. (MAΘ)
Problem 160. Express $(a^2+b^2+c^2+d^2)(e^2+f^2+g^2+h^2)$ as a sum of four squares. (Euler’s four-square identity.)
Problem 161. Show that if $x + y + z = 0$, then $x^3 + y^3 + z^3 = 3xyz$.