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

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The equation \( a^2 = b^4 + b^2 + 1 \) has been discussed regarding its integer solutions. It has been established that the only integer solutions are \( (a, b) = (\pm 1, 0) \). The discussion acknowledges this valid solution but suggests rephrasing the problem to exclude it. Ultimately, the focus remains on the existence of integer solutions, confirming that only the specified pair qualifies. The equation does not have other integer solutions beyond this pair.
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Show that the equation $a^2=b^4+b^2+1$ does not have integer solutions.
 
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anemone said:
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
 
Albert said:
$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!:o

Let me rephrase the original problem so that it does not have integer solutions, except for $(a,\,b)=(\pm1,\,0)$.
 
$a^2=b^4+b^2+1---(1)$
$\left | a \right |$ 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 :$a=\pm 1,\,\, and\,\, \, b=0$
 
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