There are two distinct integer solutions to

** **x!+8=2^{y}

a. Find both.

b. Prove no other solutions exist.

y cannot be 0, since this forces x! to be negative. This is a contradiction.

x=0 gives 2^y =9, which does not have an integer solution in y.

Therefore each of x and y is a positive integer.

Also, we note that:

6!=720 = 16*45 = M(16)

Since all x! for x>=6 is a multiple of 6, it follows that for x>=6 and y>=4 each of x! and 2^y is a multiple of 16.

Accordingly, x!=16P and 2^y= 16Q, whenever each of P and Q is a positive integer.

Substituting these values in the given equation, we have:

16P+8=16Q

=> 16(P-Q)=8

=> P-Q= 0.5,

So, lhs of the above equation is an integer but the rhs is not an integer.

This is a contradiction.

Accordingly, x<6

Then, substituting x=1,2,3,4,5 in turn, we note that:

x=4 yields, 2^y =32, giving: y=5

x= 5 yields, 2^y=128, giving: y=7

None of the other three values of x gives an integer value of y.

Consequently,

(x, y) = (4,5) and (5, 7) corresponds to all possible integer solutions to the equation under reference.

*Edited on ***June 22, 2022, 12:48 am**