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Is there an easy parametrization for the solutions of the diophantine equation $a^4+b^2=a^2+c^2$?

What I have tried is after completing the square $(a^2-\frac{1}{2})^2+b^2-c^2=\frac{1}{4}$ and replacing $ (a^2-\frac{1}{2}) \rightarrow \frac{v}{2}$, the equivalent equation is obtained $$v^2+4b^2-4c^2=1$$ which may it have a good parametrization??

Another option is using the parametrization of $x^2+y^2=z^2+w^2$, as shown in Diophantine equation $a^2+b^2=c^2+d^2$, but trying to equate $ x \rightarrow a^2,z \rightarrow a$ results in a extra diophantine equation which is difficult to parametrize further.

rgvalenciaalbornoz
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    $a^4 - a^2$ is always divisible by $4.$ Find all factorizations $a^4 - a^2 = xy$ such that $x \equiv y \pmod 2,$ then $c = \frac{x+y}{2}, ; ; ; b = \frac{x-y}{2}$ – Will Jagy Aug 11 '24 at 13:47
  • a=b=c=1 ? simplest possible? a=b=c=0 works too... – tp1 Aug 11 '24 at 20:15
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    @rgvalenciaalbornoz, One of parametric solutions is $(a,b,c)=(2n-1,\ 4n^3-4n^2+1,\ 4n^3-4n^2+2n-1).$ – Tomita Aug 12 '24 at 02:48

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As suggested in the comments, rewriting as $$a^4-a^2=c^2-b^2=(c+b)(c-b),$$ readily shows that you are parametrizing factorizations of $a^4-a^2$ into pairs of factors that are congruent mod $2$. Of course $a^4-a^2$ is divisible by $4$ for any integer $a$, and so you get all integral solutions by factoring $\tfrac{a^4-a^2}{4}=xy$ and setting $$b:=x-y,\qquad c:=x+y.$$

Servaes
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