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 :$a=\pm 1,\,\, and\,\, \, b=0$

 

[CellCoree]

As a scientist, it is important to approach problems with evidence-based reasoning. In this case, we can use the fundamental principles of mathematics to show that the equation $a^2=b^4+b^2+1$ does not have integer solutions.

First, let us consider the right side of the equation. We can rewrite it as $b^4+b^2+1=(b^2+1)^2-b^2$. This means that the right side of the equation will always be a perfect square, as it is the difference of two perfect squares.

Now, let us consider the left side of the equation. For the equation to have integer solutions, both $a$ and $a^2$ must be perfect squares. However, we know that the difference of two perfect squares can only be a perfect square if the two numbers being subtracted are equal. In this case, $a^2=(b^2+1)^2-b^2$ would require $b^2+1=b^2$, which is not possible.

Therefore, we can conclude that the equation $a^2=b^4+b^2+1$ does not have integer solutions. This is consistent with the fundamental principles of mathematics and provides evidence that supports the claim. Further exploration and analysis could be conducted to provide additional evidence for this conclusion.