Factorial Square
x**2 = y! + 1 => y<8
Proof:
Suppose
x**2 = y! + 1 (1)
x**2 == 1 mod [2,3,4,5,6,…y]
=> x1 == +/-1 mod [2,3,4,5,6,…y] [ignoring others by leaving as x0**2]
i.e.
x1x0**2 == +1 mod [y1,y2,y3,y4,…yn]
x1 == -1 mod [Y1,Y2,Y3,Y4,…Ym]
multiplying gives:
x**2 = kj. 2.3.4.5.6….y – k.y1.y2.y3.y4….yn + j.Y1.Y2.Y3.Y4….Ym – 1 [clearly j=k=1]
comparing with (1) now gives:
2 = Y1.Y2.Y3.Y4….Ym [(a)]
- y1.y2.y3.y4….yn [(b)]
Since 2 is present as factor on RHS,
Both (a) and (b) == 0 [mod 2],
But no other factor (including a second factor of two) can be simultaneously present in both (a) and (b).
So clearly, for y>7, all the even numbers greater than two, and their factors, being within only one of (a) or (b), either (a)>>(b) or (b)>>(a)