IIT JEE , AIEEE Forum - Mathematics , Algebra

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Let x² = n! - 1

x² + 1 = n!

For n=0 : x=0 ..... 1st possible n

For n=1 : x=0 ...... 2nd possible n

For n>1 : n! is even and hence x must be odd.

Let x = 2a+1

x² = 4a² + 4a + 1 = 4a(a+1) + 1

a(a+1) is an even no. , let a(a+1) = 2k

so x² = 8k + 1

x² + 1 = n!

8k+2 = n!

2(4k+1) = n!

LHS contains only one power of 2 while for n=4,5,6.... RHS contains higher powers of 2, so n=4,5,6.... are ruled out.

Checking for n=2,3 they don't make n!-1 a perfect square.

so possible solutions are n = 1,2

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