A factorial equation: Solve (1+x!)*(1+y!) = (x+y)! for natural x, y

[Stas]

Hi! Last lesson at school i learnt about factorial and on lesson we made some exercises on calculating some sums from 1 to n with factorials (for example 1*1!+2*2!+3*3!+...+n*n!)

Making my homework i found an exercise with factorials that stumped me:

For natural x and y, solve the equation (1+x!)*(1+y!)=(x+y)!

 

[pka]

Hi! Last lesson at school i learnt about factorial and on lesson we made some exercises on calculating some sums from 1 to n with factorials (for example 1*1!+2*2!+3*3!+...+n*n!)
Making my homework i found an exercise with factorials that stumped me:
For natural x and y, solve the equation (1+x!)*(1+y!)=(x+y)!

If you are really stumped, then you are over-thinking the approach.
The solution as simple as can be. Try some small numbers.

 

[Stas]

Thanks! I found out 2 small solutions (1;2) and (2;1).
Looks like this solutions are the complete answer to the exercise, but unfortunately i dont know how to prove this.

 

[pka]

Thanks! I found out 2 small solutions (1;2) and (2;1).
Looks like this solutions are the complete answer to the exercise, but unfortunately i dont know how to prove this.

\(\displaystyle (1 + x!)(1 + y!) = 1 + x! + y! + x! \cdot y!\)
Here are some facts you need think hard about.
If \(\displaystyle n\ge 5\) then the number \(\displaystyle n!\) ends in a zero (i.e. is a multiple of ten). WHY?
If \(\displaystyle (x~\&~y)\ge 5)\) then what can you say about \(\displaystyle (x+y)!,~1+x!+y!+x!\cdot y!~~?\).

 

[Stas]

\(\displaystyle (1 + x!)(1 + y!) = 1 + x! + y! + x! \cdot y!\)
Here are some facts you need think hard about.
If \(\displaystyle n\ge 5\) then the number \(\displaystyle n!\) ends in a zero (i.e. is a multiple of ten). WHY?
If \(\displaystyle (x~\&~y)\ge 5)\) then what can you say about \(\displaystyle (x+y)!,~1+x!+y!+x!\cdot y!~~?\).

Ok, i think i got this. If n>=5, n! Will always have 2 and 5 in its decomposition, so it will have a 10 in decomposition and so it will always end with 0.

Now if x and y are bigger or equal 5, (x+y)! will end in 0.
In the right part we have x!, y! and x!y! which also end on 0, but because of extra +1, the right part will end in 1, so x and y cant be >=5. So i only have to test cases like (0,4); (1,4); (2,4) ... (4,3); (4,4). Am i right?

 

[pka]

Ok, i think i got this. If n>=5, n! Will always have 2 and 5 in its decomposition, so it will have a 10 in decomposition and so it will always end with 0.
Now if x and y are bigger or equal 5, (x+y)! will end in 0.
In the right part we have x!, y! and x!y! which also end on 0, but because of extra +1, the right part will end in 1, so x and y cant be >=5. So i only have to test cases like (0,4); (1,4); (2,4) ... (4,3); (4,4). Am i right?

Yes. But you need to write the solution out clearly.

 

[Stas]

Yes. But you need to write the solution out clearly.

Ok. Thanks a lot!