Factorial vs Square
( n − 1 ) ! + 1 = n 2
Find the sum of all positive integers n such that the above equation is fulfilled.
The answer is 5.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
3 solutions
The mathematical explanation behind why n = 5 is the only possible solution can be given by the fact that the curve y = ( x − 2 ) ! grows way faster than y = x + 1 .
If you think of a smooth curve for the two functions in your mind (or you may use a graph plotter), it becomes clear that they'll intersect just at a single point which is ( 5 , 3 ) .
Log in to reply
You are absolutely correct!I also thought about the curve but forgot to mention it in my solution.Have you checked my this problem
i did it the same way ...... :)
Log in to reply
:) I think this is the easiest method please reshare and like this if you find this interesting
Testing n = 1 and n = 2 it turns out that the equality is false, so we can assume n ≥ 3 .
In particular this means two things: 1) we can divide by ( n − 1 ) that is non zero; 2) it make sense to write and consider (n-2)! and (n-3)! .
That said write
( n − 1 ) ! = n 2 − 1 = ( n − 1 ) ( n + 1 )
dividing by ( n − 1 ) we get
( n − 2 ) ! = n + 1 = n + 1 − 3 + 3 = ( n − 2 ) + 3
so shifting ( n − 2 ) to the left member
( n − 2 ) ! − ( n − 2 ) = 3
( n − 2 ) [ ( n − 3 ) ! − 1 ] = 3
We get that ( n − 2 ) divides 3 (that is a prime number). It means that ( n − 2 ) = 1 OR ( n − 2 ) = 3 .
So we have only two chances for n . Namely n = 3 or n = 5 . All other choices for n can't be solution (this is important, because we must find all the solution and sum them to get the right answer).
Just go and test the two numbers and discover that 3 fails, while 5 is a solution, and by our previous statement is the ONLY solution. So the sum of all the solution is still 5 . :)
guess the values and proove it...
you will get n=5
We have ( n − 1 ) ! + 1 = n 2
or, ( n − 1 ) ! = n 2 − 1
or, ( n − 1 ) ! = ( n + 1 ) ( n − 1 )
or, ( n − 2 ) ! = ( n + 1 ) [cancelled ( n − 1 ) in both sides,ignoring the value n = 1 ,it cannot be as 0 ! = 1 ]
Checking the first few values we can assure n can only be =5.The answer is 5 .