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If $a(x) + b(x) = x^6-1$ and $\gcd(a(x),b(x))=x+1$ then find a pair of polynomials of $a(x)$,$b(x)$.

Prove or disprove, if there exists more than 1 more distinct values of the polynomials.

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Hint: What do you get if you divide $x^6-1$ by $x+1$? –  Aleks Vlasev Oct 16 '12 at 6:35
    
I've removed algebra tag, since we don't use algebra tag anymore, see meta for details. –  Martin Sleziak Oct 19 '12 at 13:56

3 Answers 3

up vote 2 down vote accepted

There is too much freedom. Let $a(x)=x+1$ and $b(x)=(x^6-1)-(x+1)$. Or else use $a(x)=k(x+1)$, $b(x)=x^6-1-k(x+1)$, where $k$ is any non-zero integer.

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Thank you.I don't what i was thinking :p –  Sai Krishna Deep Oct 16 '12 at 7:08

Since $\gcd\{a(x),b(x)\}=x+1$, there are polynomials $p(x)$ and $q(x)$ such that $a(x)=(x+1)p(x)$ and $b(x)=(x+1)q(x)$.

$$x^6-1=(x+1)\left(x^5-x^4+x^3-x^2+x-1\right)\;,$$

so

$$p(x)+q(x)=x^5-x^4+x^3-x^2+x-1\;.\tag{1}$$

You can easily find several pairs of relatively prime polynomials $p(x)$ and $q(x)$ that satisfy $(1)$, and they give you pairs $a(x),b(x)$.

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You just have to make sure that $\gcd(p,q)=1$. For example, let $p(x)=x^k$. –  Hagen von Eitzen Oct 16 '12 at 8:35

Hint $\ \ $ If $\rm\:a\!+\!b = c\:(x\!+\!1)\:$ then

$$\rm\:x\!+\!1 = (a,b) = (a,c\,(x\!+\!1)\!-\!a) = (a,c\,(x\!+\!1))\iff a = d(x\!+\!1),\,\ (d,c) = 1$$

Thus for $\rm\:c(x\!+\!1) = x^6\!-\!1\:$ it's true $\rm\iff a = d(x\!+\!1),\:$ for $\rm\,d\,$ coprime to $\rm\:c = (x^6\!-\!1)/(x+1),\:$ or, equivalently, $\rm\,d\,$ coprime to $\rm\:x\!-\!1,\ x^2\pm x + 1$

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