I think the problem is wrong $(x+1)! \ge x!$ so it may be $+2^y$

Ignore this

If this problem is true, this is way too easy for the national olympiad second round The problem changed Ignore this

Seungjun_Lee wrote: I think the problem is wrong $(x+1)! \ge x!$ so it may be $+2^y$ Sorry, it was a typo

Then $x\cdot x!=2^y$.When $x\geqslant 3$,$3\mid x!$.So $x\leqslant 2$. $(x,y)=(2,2),(1,0)$.

$2^y = x \times x!$ Hence, $x! = 2^k$ If $x \ge 3; \,\, 3|2^k$ contradiction This means $x = 0\,or\,1\,or\,2$ $x \neq 0$ $x=1$ then $y=0$ $x=2$ then $y=2$ $(x,y) = (1,2) or (2,2)$

We have that $2^y=x\cdot x!$. Therefore $x$ and $x!$ are powers of two. If $x\geq 3$, then $3\mid x!$ and $x!$ is not a power of two. Therefore $x\leq 2$. If $x=0$, then $1+2^y=1$, but there is no solution. If $x=1$, then $1+2^y=2$ and $y=0$. If $x=2$, then $2+2^y=6$ and $y=2$. Therefore, the only satisfying couples are $(1, 0)$ and $(2, 2)$.

Note, that $(x+1)!-x!=(x+1)x!-x!=x\cdot x!=2^y.$ Hence, $x, x!$, are powers of $2,$ if $x \ge 3$, then $x!$, is not a power of $2.$ If $x=0$, no solution, if $x=1$, we have $y=0,$ if $x=2, y=2$ hence, our solutions are $\boxed{(1,0),(2,2)}.$

We rearrange to obtain, \[2^y = (x+1)!-x!=xx!\]Now, note that if $x\geq 3$, then, $3 \mid xx!$ so we also have that $3 \mid 2^y$ which is a clear contradiction. Thus, we have a few possibilities to check. When $x=0$, $2^y=0$ which has no solutions. When, $x=1$ we have that \[2^y=2!-1!=1\]so $y=0$. When, $x=2$, we have that \[2^y=3!-2!=4\]so $y=2$. Thus, all possible pairs of non-negative integer solutions $(x,y)$ which satisfy the required equation are, \[(x,y)=(1,0),(2,2)\]

The expression can be re-written as \[2^y=x\cdot x!\]If $x\ge 3$, then $3\mid x\cdot x!=2^y$ which is a contradiction. Also, \[x=0\implies 2^y=0\text{ which is another contradiction}\]So, $x=1,2$ gives $\boxed{(x,y)=(1,0),(2,2)}$.

sman96 wrote: Find all possible non-negative integer solution $(x,y)$ of the following equation- $$x! + 2^y =(x+1)!$$Note: $x!=x \cdot (x-1)!$ and $0!=1$. For example, $5! = 5\times 4\times 3\times 2\times 1 = 120$. We get that $2^y=x\cdot x!$, so $x!$ must be a power of $2$. Clearly if $x\ge3$ then it cannot be the case as we will have $3\mid x!$ so checking $x=1,2$ we get solution pairs $(x,y)=(1,0),(2,2)$.

Let's manipulate the equation: $$x! + 2^y =(x+1)!$$$$2^y =(x+1)!-x!$$$$2^y =x.x!$$Notice that L.H.S only contains factor of 2 (or 1), if $x!$ is greater than 2, than it has prime factor other than 2, contradiction. Casework tells us that the only solutions are $(x,y)=(2,2),(1,0)$.

we claim that $(x,y)=(1,0);(2,2)$ For $x>2$ , $(x+1)!-x!$ is divisible by $3$ , but $2^y$ is $1$ or $2$ modulo $3$ , a contradiction.

Clearly $2^y = x \cdot x!$. Now if $x \ge 3$, we have $3 \mid x \cdot x!$ but 3 does not divide $2^y$, contradiction. Therefore $x \le 2$. $x = 0$ gives us $2^y = 0$ which has no solution. $x = 1$ gives us $2^y = 1 \implies y = 0$, giving the solution $(1, 0)$. $x = 2$ gives us $2^y = 4 \implies y = 2$, giving the solution $(2, 2)$. Therefore the only solutions are $(x, y) = (1, 0), (2, 2)$.

Here, $x!+2^y=(x+1)!$ can be written as $2^y=x!\cdot x$. As both sides of the equation need to be divisible by $2$, $x$ can't be more than $2$. Therefore, $x$ needs to be equal to $0$, $1$ or $2$. If $x=0$, then $2^y=0$, which has no solution. If $x=1$, then $2^y=1$, which gives us $y=0$. If $x=2$, then $2^y=4$, which gives us $y=2$. Therefore, the possible solutions are $(1,0)$ and $(2,2)$.

if we take mod 10 we get that x<5 because if x>4 then the number get 0 from mod 10 , but $2^y$ isn't get 0 from mod 10 , let's check x values in x<5 , there are (1;0) and (2,2) solutions.

$(x+1)!-x!=2^y, x!*x=2^y$ So, x! should not have any primes factors except for 2. It can only be satisfied when $ x = 1, 2 $ ($x \neq 0$ because $2^y \neq 0$) $ x = 1 $ then $y =0 $ $ x = 2 $ then $y = 2 $ $\therefore (x,y) = (1, 0), (2,2)$