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Thread: factorising and proving

  1. #1
    shosho
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    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

    thanx in advance
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  2. #2
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    Hello, shosho!

    Here are the factoring problems .. .


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

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

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



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

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

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

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


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

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

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

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  3. #3
    Super Member

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    Hello again, shosho!

    The second problem requires some divisibility theorems.

    For a positive integer $\displaystyle n$:

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

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


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


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

    Hence, $\displaystyle N$ is divislble by 3.


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

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

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

    Hence, $\displaystyle N$ is divisible by 29.


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

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