IIT JEE , AIEEE Forum - Mathematics , Algebra

hsbhatt

Find all n such that n!-1 is a perfect square

nadeemoidu

All squares are of the form 4k or 4k + 1.

So the only possible values are n=1 and n=2

iitkgp_bipin

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

pantpranav

n is perfect square if n=4x or n=4x+1 where x is a whole number.
So the only possible solutions are 1 & 2

hsbhatt

or you can use that any perfect square is of the form 3k or 3k+1. you get the same result.

unspecified

n=1,2

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