Thread: factorising and proving

1. factorising and proving

hey im having some trouble with these two problems ..any help would be great

Factor:
1a) x^8 + 2x^4y^4 + 9y^8
b) a^4 +b^4 + c^2 - 2(a^2b^2 + a^2c + b^2c)

2) prove that

20^22 - 17^22 +4^33 -1

is divisible by 174

2. Hello, shosho!

Here are the factoring problems .. .

1a) Factor: . $x^8 + 2x^4y^4 + 9y^8$
$\text{Add and subtract }4x^4y^4\!:\;\;x^8 \:+ \:2x^4y^4 \:\underbrace{+\, 4x^4y^4} \:+ \;9y^8 \:\underbrace{-\, 4x^4y^4}$

$\text{We have:}\;\;\left(x^8 + 6x^4y^4 + 9y^8\right) - \left(4x^4y^4\right) \;=\;\underbrace{\left(x^4 +3y^4\right)^2 - \left(2x^2y^2\right)^2}_{\text{difference of squares!}}$

Therefore: . $\left(x^4 + 3y^4 - 2x^2y^2\right)\left(x^4+3y^4 + 2x^2y^2\right) \;\;=\;\;\boxed{\left(x^4 - 2x^2y^2 + 3y^4\right)\left(x^4 + 2x^2y^2 + 3y^4\right) }$

1b) . $a^4 + b^4 + c^2 - 2(a^2b^2 + a^2c + b^2)$
You won't like this one . . . I promise!

We have: . $a^4+ b^4 + c^2 - 2a^2b^2 - 2a^2c - 2b^2c$

$\text{Add and subtract }4a^2b^2\!:\;\;a^4 + b^4+ c^2 - 2a^2b^2\: \underbrace{+\, 4a^2b^2}\: -\: 2a^2c \:- \:2b^2c\: \underbrace{-\,4a^2b^2}$

$\text{We have: }\;(a^4 + b^4 +c^2 +2a^2b^2 - 2a^2c - 2b^2c) \:-\: (4a^2b^2) \;\;= \;\;\underbrace{(a^2 + b^2 - c)^2 - (2ab)^2}_{\text{difference of squares!}}$

Therefore: . $(a^2 + b^2 - c - 2ab)\!\cdot\!(a^2 + b^2 - c + 2ab)$

. . . . . . $= \;(a^2-2ab+b^2 - c)\!\cdot\!(a^2+ 2ab+b^2 - c)$

. . . . . . $= \;\left[(a-b)^2 -c\right]\!\cdot\!\left[(a+b)^2 - c\right]$ . . . . Wow!

3. Hello again, shosho!

The second problem requires some divisibility theorems.

For a positive integer $n$:

. . $[1]\;\;a^n - b^n$ is always divisible by $(a - b).$

. . $[2]\;\;a^n +b^n$ is divisible by $(a + b)$ for odd $n.$

2) Prove that: . $N \;=\;20^{22} - 17^{22} +4^{33} -1$ is divisible by 174.
$\text{We have: }\;N \;=\;\underbrace{(20^{22} + 4^{33})}_{\text{even}} -\underbrace{(17^{22} + 1)}_{\text{even}} \;=\;\text{even}$
Hence, $N$ is divisible by 2.

From [1]: $20^{22} - 17^{22}$ is divisible by $(20-17) = 3$
. . . .and $4^{33} - 1^{33}$ is divisible by $(4-1) =3$

Hence, $N$ is divislble by 3.

We have: . $N \;=\;(20^{22} + 4^{33}) -(17^{22} + 1)$

$20^{22} + 4^{33} \;=\;(20^2)^{11} + (4^3)^{11} \;=\;400^{11} +64^{11}$
. . From[2], this is divisible by: $400 + 64 \:=\:464 \:=\:16\cdot29$

$17^{22} +1^{22} \;=\;(17^2)^{11} + (1^2)^{11} \;=\;289^{11} + 1^{11}$
. . From [2], this is divisible by $289 + 1 \:=\:290 \:=\:10\cdot29$

Hence, $N$ is divisible by 29.

Since $N$ is divisible by 2, 3 and 29, $N$ is divisible by: $2\!\cdot\!3\!\cdot\!29 \:=\:174$