Can I find solutions to $a^4 + a^2 + a = b^2 + b$, $a,b \in \mathbb{Z}$ and $ 1 < a < b$?

Question

I was wondering if anyone could point me in the correct direction for either finding a solution to my problem or proving that it does not exist.

$$a^4 + a^2 + a = b^2 + b \;\text{ for }\; a,b \in \mathbb{Z} \;\text{ and }\; 1 < a < b$$

To be honest I don't really believe there is a solution (my brute force program yielded no solutions). But how do I prove this in the general sense? If you don't have time to provide a proof a nod in the right direction would be greatly appreciated.

Edit: Like some people pointed out, I make a mistake in the initial question, the constraints on a and b should be: $1 < a < b$ instead of $1 < b < a$. I have fixed the question to reflect this change.

Answer 1

We can write the equation as $$a=(b-a^2)(b+a^2+1).$$ But if $b$ is positive the right factor $b+a^2+1$ is strictly greater than $a$, so equality can never be attained.

Answer 2

Hint: $a^4+a^2+a > a^2+a > b^2+b$ if $a > b> 1$

Answer 3

There is no solution because $$a^4+a^2+a>b^2+b$$ Indeed, $$a^2+a>b^2+b$$

Answer 4

Solution: $a > 1, b = \frac{1}2\left(\sqrt{4a^4+4a^2+4a+1}-1\right)$.