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.

Here is the statement: Let $a,b \in \mathbb{Z}$ positive integers such that $a^2=b^4+b^3+b^2+b+1.$ Prove $b=3.$

I've tried is the following: Let $\Sigma=b^4+b^3+b^2+b+1$. If $a\equiv 0\mod 3$, then $a^2\equiv 0 \mod 3$. If $a\equiv 1\mod 3$ or $a\equiv 2\mod 3$, then $a^2\equiv 1\mod 3$. In the other hand, $b\equiv 0\mod 3$ or $b\equiv 2\mod 3$, then $\Sigma\equiv 1\mod 3$ and if $b\equiv 1\mod 3$, $\Sigma\equiv 1\mod 3$. Thus $a=3n$ or $a=3n+2$ for some $n\in\mathbb{Z}$, and $b=3m$ or $b=3m+2$ for some $m\in\mathbb{Z}$. In a similar way, I have prove $a$ cannot be even.

If $a=3n$, for some $n\in \mathbb{Z}$ and $b=3m$, for some $n\in \mathbb{Z}$, we arrive at the following equation $(3n-1)(3n+1)=3m(3³m³+3²m²+3m~1).$ So, $a\equiv 0\mod 3$ or $b\equiv 0\mod 3$ but not both at same time.

I think I should use the Legendre symbol. In this case $(\frac{\Sigma}{3})=0 \text{ or }1$.

share|improve this question
1  
Welcome to math.SE! Unfortunately your question does not show that you have thought about the problem. Please explain what you've tried so far, and where you are stuck. –  Zev Chonoles Dec 14 '11 at 1:52
    
Thank you for your feedback. –  Juliho Castillo Dec 14 '11 at 1:55
    
On this site, it is not considered polite to post questions when they are in an imperative tone. It is perfectly fine if you are not even sure where to start - but please say so. Also, if this is homework, there is a "homework" tag that should be added. –  Zev Chonoles Dec 14 '11 at 1:58
1  
Now you know that if $a \equiv 0 (mod 3)$ then $b$ is not divisible by 3. Thus, $b \equiv 1 (mod 3)$ or $b \equiv 2 (mod 3)$. If $b \equiv 1 (mod 3)$ then $b^4 + b^3 + b^2 + b + 1 \equiv 2 (mod 3)$ which is not possible since $a^2 \equiv 0 (mod 3)$. Similarly, if $b \equiv 2 (mod 3)$ then $b^4 + b^3 + b^2 + b + 1 \equiv 1 (mod 3)$ which is not possible. Thus, we cannot have $a \equiv 0 (mod 3)$. –  Rankeya Dec 14 '11 at 2:39
add comment

1 Answer

If $b$ is even then $b^4+b^3+b^2+b+1$ is caught between $$(b^2+(b/2))^2=b^4+b^3+(1/4)b^2$$ and $$b^4+b^3+(9/4)b^2+b+1=(b^2+(b/2)+1)^2$$

If $b$ is odd, have a look at $(b^2+(b-1)/2)^2$ and $(b^2+(b+1)/2)^2$.

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.