A Nice Equation

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[benjaminvatovez]

1) No condition of existence
2) 1-x^8=2^(x²)>0 thus -1<x<1
3) If x is a solution then -x is a solution
4) Let us consider 0<=x<1. As f(x)=2^(x²)+x^8-1 strictly increases on this interval as the sum of two increasing functions, there is at most one x such that f(x)=0.
5) x=0 is a solution
Conclusion: x=0 is the only solution on [0, 1) and thus on (-1, 0] also.

[renesperb]

If you notice that 1 - x^8 has the maximum 1 and e^x^2 has the minmum 1, it is clear that x=0 is the only real solution.

[cdiesch]

If you admit complex solutions there is two other solutions.
I only don‘t know how to calculate. But if isubstitute y=x^2, the graph shows me 2 solutions:
y=0 or y≈ - 0.80509
so
x=0 or x≈ 0.899727 i or x≈-0.899727 i

[pavlopanasiuk]

That's too much job over this example. Set t=x^2, t is therefore non-negative. 2^t >= 2^0 = 1 (power function grows); 1-t^4 <= 1 (t^4 grows). The intersection is resolved at 2^t = 1 -> t=0.