Find all pairs (n, t) of positive integers that satisfy the equation:
(n+1)t - 1 = n!
Provide adequate reasoning for your answer.
First, the case where t=1. Then the equation reduces to n=n!. Trivially 1!=1 and 2!=2.
Then the first two solutions are (n,t)=(1,1) and (2,1).
If n=3 then 4^t - 1 = 6. This has no integer solution for t.
If n=4 then 5^t - 1 = 24. This makes t=2. Then (n,t)=(4,2) is another solution.
Then we have the "large" values where t>=2 and n>=5.
Assume t>=n. Then (n+1)^t - 1 > n^t > n^n > n!. So then there are no solutions with t>=n.
Now lets take the equation (n+1)^t - 1 = n! mod n^2.
By using binomial expansion the left side equals n*t mod n^2. I will then split this into two cases: "n>=6 and composite" and "n>=5 and prime".
"n>=6 and composite"
Wilson's theorem tells us that in this case (n-1)! = 0 mod n. Then n! = 0 mod n^2. n*t cannot equal zero or be large enough to get to n^2 with our constraints, so no solutions in this case.
Wilson's theorem tells us that in this case (n-1)! = n-1 mod n. Then n! = n^2-n mod n^2. This is still a multiple of n and then this needs to equal n*t, so t=n-1 is necessary.
Then (n+1)^(n-1) - 1 = n! But (n+1)^(n-1) > n^(n-1) > n!, so no solutions in this case either.
Then that leaves us with only the three solutions (n,t)=(1,1), (2,1) and (4,2).
Edited on April 4, 2023, 10:20 am
Analytic Solution
