Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

3 Answers

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.

share|improve this answer
    
Thank you.I don't what i was thinking :p –  Sai Krishna Deep Oct 16 '12 at 7:08
add comment

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 $a(x),b(x)$.

share|improve this answer
    
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
add comment

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 $\rm\:x\!-\!1,\ x^2\pm x + 1$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.