2022 IMO Problems/Problem 5
Problem
Find all triples of positive integers with prime and
Video solution
https://www.youtube.com/watch?v=-AII0ldyDww [Video contains solutions to all day 2 problems]
Solution
Case 1:
- Since is indivisible by , then must also be indivisible by .
- If , then is divisible by , so must be a divisor of , but obviously has no solutions and we ruled out already. For , let's show that there are no solutions using simple inequalities.
- If , then and by throwing away the remaining (non-negative) terms of binomial theorem. For any solution, , which is impossible for . That leaves us with and , but for any integer (proof by induction), so there are no solutions.
- If , RHS is at most and LHS is at least (again from binomial theorem), which gives no solutions as well.
Case 2:
- Since is divisible by , then must also be divisible by .
- In addition, RHS is at most , so . We may write , where .
- Since is a divisor of and it must also be a divisor of , so and . We're looking for solutions of .
- Let's factorise : if with odd and , it's
- Since and contain an odd number of odd terms (remember the assumption aka ), they're odd. Also, modulo , so and each following term is even but indivisible by . The highest power of dividing is therefore where is the highest power dividing .
- In comparison, has factors , etc (up to ), and other even factors, so it's divisible at least by . Since for , the only possible solutions have or .
- If , we reuse the inequalities and to show that there are no solutions for .
- Finally, isn't a factorial, and .
Case 3:
Just like in case 2, is divisible by so must also be divisible by . However, and are also both divisible by , so remainders modulo tell us that no solutions exist.
Conclusion:
The only solutions are .
Solution 2
I considered the cases:
1) If \(a\) is even, then it must:
1.a) \(b \neq 1\) and \(p > 2\) 1.b) \(b!\) is even and \(p = 2\)
2) If \(a\) is odd, then it must:
2.a) \(b \neq 1\) and \(p = 2\) 2.b) \(b!\) is even and \(p > 2\)
Examining 1.a), we end up with an equation of the form \(a^p = p+1\), which has no integer solutions.
In case 1.b), \(b!\) can take values: 2, 6, 24, 120, 720, ..., so \(b!+2\) takes values: 4, 8, 26, 122, 722, .... We observe that the only perfect square is 4 among the possible cases, as for \(b \geq 5\), the result ends in 2, which is not a perfect square. Therefore, we have the triple \((2,2,2)\).
Case 2.a) yields \(a^2 = 3\), which is rejected.
Examining the last case 2.b), we have for \(b!\) the values: 2, 6, 24, 120, 720, 5040,
i) \(b! = 2\), then \(a^p = 2+p\), which is rejected. ii) \(b! = 6\), then \(a^p = 6+p\), also does not give integer solutions. iii) \(b! = 24\), then \(a^p = 24+p\), giving a solution for \(p=3\). Thus, we have the triple \((3,4,3)\). iv) \(b \geq 5\): \(a^3 = b! + 3\), so \(a\) must end in 3 or 7, which does not give solutions. \(a^5 = b! + 5\), so \(a\) must end in 5, also not giving solutions. \(a^7 = b! + 7\), so \(a\) must end in 3, again not giving solutions. \(a^{11} = b! + 11\), no \(a \in \mathbb{Z^+}\) satisfies this. \(a^{19} = b! + 19\), similarly, no solutions. Other prime exponents reduce to these, as their last digit is one of \(\{1,3,7,9\}\)
The analysis of the last case is incomplete, which is why I wasn't initially sure about the number of triples. Therefore, with this approach (which is not strictly documented), we find the triples: \((2,2,2), (3,4,3)\).
See Also
2022 IMO (Problems) • Resources | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All IMO Problems and Solutions |