Does the equation $a^2=b^4+b^2+1$ have integer solutions?

[anemone]

Show that the equation $a^2=b^4+b^2+1$ does not have integer solutions.

 

[Albert1]

Show that the equation $a^2=b^4+b^2+1$ does not have integer solutions.

$a=\pm 1 ,\, and \,\, b=0 $ are two sets of integer solutions

 

[anemone]

$a=\pm 1 ,\, and \,\, b=0 $ are two sets of integer solutions

Oh...it was my bad that I didn't see $(a,\,b)=(\pm1,\,0)$ is the valid solution to this problem. Sorry!

Let me rephrase the original problem so that it does not have integer solutions, except for $(a,\,b)=(\pm1,\,0)$.

 

[Albert1]

Spoiler

$a^2=b^4+b^2+1---(1)$
$\left | a \right |$ must be odd
let $a=2x+1,\,\, and, \,\, y=b^2\geq 0$
from (1) we have:
$4x^2+4x+1=y^2+y+1$
$\therefore 4x(x+1)=y(y+1)---(2)$
the only possible solutions for (2) will be :$4x(x+1)=y(y+1)=0$
we get $x=0 ,\,\, or,\,\, x=-1$
and the corresponding solutions of $a,\,\, and \,\,\, b $ will be :$a=\pm 1,\,\, and\,\, \, b=0$