Factorials and squares...

[The legend]

1. The problem statement, all variables and given/known data

For some natural N the no of positive integers x satisfying the equation
1! + 2! + 3! + ........ + x! = (N)2 is:
A)one
B)two
C)infinite
D)none


3. The attempt at a solution
I have no idea of how to start.. never came across such problems.
By trial i have got two values for x being 1 and 3. And the answer is also two . But how do i prove it?

Please help.(just a starting guidance would be very useful)

[fzero]

Past x=3, the sum of factorials always has a 3 in the last place, s4=33, s5 = 153, ... You need to argue that the square of a natural number cannot end with a 3.

[JonF]

Or better yet argue n(n-1)(n-2) > n^2 for n>3. If you can prove this you know for N>3 your relation can't hold, since n! > n^2 will be true.

[The legend]

Past x=3, the sum of factorials always has a 3 in the last place, s4=33, s5 = 153, ... You need to argue that the square of a natural number cannot end with a 3.

I found this easier, Thanks. Though i could find the answer by your method too JonF thanks!

[JonF]

You need to prove this fact if you use his method, “Past x=3, the sum of factorials always has a 3 in the last place"

n(n-1)(n-2) > n^2 for n>3 Is simple algebra.

If you expand n(n-1)(n-2) you get (n^2-n)(n-2) = n^3 – 2n^2 –n^2 +2n= n^3 – 3n^2 +2n = n^2(n-3) +2n

n^2(n-3) +2n is clearly positive if n>3.

[The legend]

Ok.. got it.

[sjb-2812]

Or better yet argue n(n-1)(n-2) > n^2 for n>3. If you can prove this you know for N>3 your relation can't hold, since n! > n^2 will be true.

I'm not overly sure this is useful, as the way I read the original question x and N are two (possibly) different numbers.

It could be, for instance that 1! + 2! + 3! + 4! + 5! == 162, instead of the 52 you seem to be suggesting. Of course it isn't, but have I misinterpreted what you're saying?