Call a positive integer emphatic if it can be written in the form a2+b!, where a and b are positive integers. Prove that there are infinitely many positive integers n such that n, n+1, and n+2 are all emphatic. Allen Wang
Problem
Source: ELMO Shortlist 2024/N2
Tags: Elmo, number theory
22.06.2024 19:04
Consider a=b!4−1. Then, a2+b!, (a+2)2+1!, (a+2)2+2! works.
22.06.2024 19:09
Not a solution but related: By \mod 8 and finite case check for small values, you can prove that at maximum 4 consecutive numbers are emphatic. I don't know if there are infinitely n with n,n+1,n+2,n+3 emphatic, but n=168 works.
22.06.2024 19:14
It suffices to show that there are infinitely many solutions to a^2 + b! = m^2, as 1 = 1! and 2! = 2. We want (m-a)(m+a) = b!, so take b\geq 4 and we can have m-a = 2, m+a = \frac{b!}{2}, i.e. a = \frac{b!}{4} - 1, m = \frac{b!}{4} + 1.
23.06.2024 14:55
IAmTheHazard wrote: Call a positive integer emphatic if it can be written in the form a^2+b!, where a and b are positive integers. Prove that there are infinitely many positive integers n such that n, n+1, and n+2 are all emphatic. Allen Wang x^2+1=a^2+1!,x^2+2=a^2+2! So we need to prove that there is infinite solution to the equation x^2+3=a^2+b! For x=a+c we get c(c+2a)=b!-3 And for b>=3 we can get c=3 and we are done.
11.07.2024 05:51
Note that a^2 + 1, a^2 + 2 are clearly emphatic for any positive integer a. Claim: There are infinitely many perfect square that are emphatic. Proof: For any b > 2, we note that if b! = 4k, then (k+1)^2 = (k-1)^2 + b!, so (k+1)^2 is emphatic. There are clearly infinitely many possible values of k, so the claim is proven. \square Now, there exist infinitely many a where a^2 is emphatic, and a^2 + 1, a^2 + 2 are also emphatic, so taking n = a for any such a works.
29.08.2024 18:57
Let d be a positive divisor of \frac{m!}{4} where m\geq 4 is an integer. Let n=(\frac{m!}{4d}+d)^2. Then the following equalities hold. n=(\frac{m!}{4d}-d)^2+m! n+1=(\frac{m!}{4d}+d)^2+1!n+2=(\frac{m!}{4d}+d)^2+2! Therefore n is emphatic. Changing values of m and d gives us infinitely many emphatic positive integers. Therefore there exists infinitely many emphatic positive integers.
18.10.2024 18:55
Very Easy. Take b such 2|b and a=\frac{(b-1)!-b}{2} And take n=a^2+b!. Note that n is a square number. n+1=n+1! and n+2=n+2! so there exists infinite triplets (n,n+1,n+2) \square