A factorial product

Parity arguments are enough here.

Let's assume $x \le y$ . If $x>1$ , the left-hand side of the expression is odd, whereas the right-hand side is even; so we must have $x=1$ .

Now $2(1+y!)=(1+y)!$ . If $y=1$ , the left-hand side is $4$ and the right-hand side is $2$ , so $(1,1)$ is not a solution.

If $y=2$ , we get $6$ on both sides, so $(1,2)$ is a solution.

If $y \ge 3$ then the right-hand side is a multiple of $4$ , but the left-hand side is not; so there are no further solutions. This gives the answer $1+2=\boxed3$ .

If $x+y \ge 5$ , then the unit’s digit of $(x+y)!$ shall be zero. To fulfil that, we must have $x, y < 5$ .

This is because if $x,y > 5$ , then $x!, y!$ would have $0$ at their unit’s place and the LHS would have $1$ as the unit’s digit.

We check and see there are no possibilities.

The rest remains to check $x+ y < 5$ which gives us the required solution $(x, y) = (1, 2), (2, 1)$ .

if x is the smaller of x and y.Simply taking mod x! solves the problem.

(1+x!)(1+y!)=x!+y!+x!y!+1 = (x+y)!... Now,x!y!+1=(x+y)!-(x!+y!)....which implies

x!y!=x+y-1.......... But if x and y get too big then this is impossible... Now taking x=1....... We get y!=y....... So y=2....... x+y=3

We will find some $x,y\in \mathbb{N}$ such that $\left( 1 + x! \right) \left( 1 + y! \right) = \left( x + y \right)!$ If $x + y \le 1$ then one of $x,y$ is 1 while the other is 0 or both $x\text{ and } y$ are zero . In any case the left hand side is $4$ and the right hand side is $1$ thus this is impossible and so $x + y > 1 \Leftrightarrow x + y \ge 2$

Though if $x + y\ge 2$ then we observe that $\left( x + y \right) !$ is even, the left hand side is a product of two integers and so one of them must be even. without loss of generality we consider $\left( 1 + x! \right)$ even, then we know $x!$ must be odd this is only true if $x= 1$ or $x= 0$ let's try $x= 1$ .

If $x= 1$ then we know that $x!= 1$ and then our original equation becomes $2\left( 1 + y! \right) = \left( 1 + y \right) !$ That is asking 2 multiplied by what number is a factorial, after some thought we know that number to be 3 and so $1 + y! = 3 \Leftrightarrow y= 2$ , and we see that yes $2\left( 3 \right) = 3!$ and so we know $x= 1\text{ and } y= 2$

If x > 1 and y > 1, then x! + 1 is odd, y! + 1 is odd, so the left side is odd, whereas (x + y)! is even, an impossible situation. Therefore, at least one of x,y must equal 1. Suppose x = 1. Then 2 (1 + y!) = (1 + y)!. When y is 3 or more, clearly the left side is less than the right side, so y = 1 or y = 2. If y = 1, 2 2 = 3! = 6, impossible, so y = 2 and 2*3 = 3! = 6. So the answer is 1 + 2 = 3.

If we can show that the equation can only be true for small numbers, then trial and error will be a viable method.

The factorial function $n!$ tends to increase very quickly as you increase $n$ --much quicker than simply multiplying two numbers together, because by definition a factorial multiplies many numbers together! Therefore, we can at least intuit that for large numbers the factorial of the sum of two numbers should be much greater than the product of their individual factorials. That is,

$x!y!<<(x+y)!$ .

In addition, for large numbers, $x!$ is much greater than 1, so adding 1 each to $x!$ and (y!) on the left-hand side will not make a big difference. We can therefore also say that, for large numbers,

$(1+x!)(1+y!)<<(x+y)!$ .

Therefore, the equation in question can only be true for some small numbers $x$ and $y$ . If we insert 0 and 0 for x and y, we get a difference of 3 between the left- and right-hand sides of the equation. Not too bad! This means we shouldn't have to go far to get the answer. Eventually, we'll get both sides equal when x and y and 1 and 2 (not necessarily respectively).