The given expression is

${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{f(x)}& =& (1+(a+x{)}^2)(1+(a-x{)}^2)\\ \multicolumn{1}{c}{}& =& (1+x^2+a^2+2\mathrm{ax})(1+x^2+a^2-2\mathrm{ax})\\ \multicolumn{1}{c}{}& =& (1+x^2+a^2{)}^2-4a^2{x}^2\\ \multicolumn{1}{c}{}& =& x^4+2x^2(1+a^2)+a^4-4a^2{x}^2\\ \multicolumn{1}{c}{}& =& x^4+2x^2(1-a^2)+(1+a^2{)}^2\end{array}}$

Note that this is a quartic in $x$ with the coefficient of $x^4$ positive and so we should have expected one or three turning points.

Differentiating $f$ to find the minima:

${\displaystyle f\text{'}(x)=4x^3+4x(1-a^2)=4x(x^2+(1-a^2))}$

Case 1: $(1-a^2)>0$

The derivative $f\text{'}(x)=0$ if and only if $x=0$ and the second derivative $f"(x)=12x^2+4(1-a^2)>0$. So there is one minimum value $f(x)=(1+a^2{)}^2$ where the position of X, at $x=0$, is independent of $a$.

Case 2: $(1-a^2)<0$ The derivative $f\text{'}(x)=0$ for $x=0$ and $x=\pm \sqrt{(}{a}^2-1)$.

The second derivative $f"(x)=12x^2+4(1-a^2)<0$ at $x=0$ which now gives a maximum value but $f"(x)=12x^2+4(1-a^2)>0$ for $x=\pm \sqrt{(}{a}^2-1)$ giving two minimum points on the quartic where $x=\pm \sqrt{(}{a}^2-1)$. The minimum value at each point is $f(x)=4a^2$ where the position of X for these minimum values is clearly dependent of $a$.

Case 3: $(1-a^2)=0$ Note that where $a^2=1$ there is a single minimum $f(x)=4$ at $x=0$ giving continuity between Case 1 and Case 2.

The most likely first conjecture agrees with Case 1 but does not allow the possibility of Case 2. Realising that the function we are minimizing is a quartic we should have taken into account the possibility of two minimum values.