For some natural number $N$,the number of positive integers $'x'$ satisfying the equation $1!+2!+3!+...+(x!)=(N)^2$
Please give some hints.
2026-03-31 06:33:36.1774938816
For some natural number $N$,the number of positive integral $'x'$ satisfying the equation is?
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Clearly $(x,N)= (1,1)$ and $(3,3)$ satisfy the equation.
Now up until $x=4$, the LHS is $1+2+6+24=33$ and after $x=4$, $k!=10l$ is a multiple of $10$. Therefore for all $x$ greater than equal to $5$ the sum is of the form $10p+3$ which can never be perfect square. Therefore only 2 ordered pairs exist.