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The number of values positive integers $$n$$ for which $$1!+2!+\cdots+n!$$ is a perfect square is
For $$n=1$$ and $$n=3$$, the sums are $$1$$ and $$9$$, which are perfect squares, while the sums for $$n=2$$ and $$n=4$$ are $$3$$ and $$33$$. For every $$n\geq5$$, all later factorials are divisible by $$120$$, so the sum is congruent to $$33\pmod{120}$$. Since $$33$$ is not a quadratic residue modulo $$120$$, no further values work, giving exactly $$2$$ values.
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