# A quartic diophantine equation

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 $(\frac{\Sigma}{3})=0 \text{ or }1$.

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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

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 $(b^2+(b+1)/2)^2$.