Maths Olympiad Prep

Library / /102 of 740

, 2015

Number theory Difficulty 4.6 AIME Find the answer United States

Problem:
Consider the following seven false conjectures with absurdly high counterexamples. Pick any subset of them, and list their labels in order of their smallest counterexample (the smallest nn for which the conjecture is false) from smallest to largest. For example, if you believe that the below list is already ordered by counterexample size, you should write "PECRSGA".
- P. (Polya's conjecture) For any integer nn, at least half of the natural numbers below nn have an odd number of prime factors.
- E. (Euler's conjecture) There is no perfect cube nn that can be written as the sum of three positive cubes.
- C. (Cyclotomic) The polynomial with minimal degree whose roots are the primitive nnth roots of unity has all coefficients equal to 1,0-1,0, or 1.
- R. (Prime race) For any integer nn, there are more primes below nn equal to 2(mod3)2(\bmod 3) than there are equal to 1(mod3)1(\bmod 3).
- S. (Seventeen conjecture) For any integer nn, n17+9n^{17}+9 and (n+1)17+9(n+1)^{17}+9 are relatively prime.
- G. (Goldbach's (other) conjecture) Any odd composite integer nn can be written as the sum of a prime and twice a square.
- A. (Average square) Let a1=1a_{1}=1 and ak+1=1+a12+a22++ak2ka_{k+1}=\frac{1+a_{1}^{2}+a_{2}^{2}+\ldots+a_{k}^{2}}{k}. Then ana_{n} is an integer for any nn.
If your answer is a list of 4n74 \leq n \leq 7 labels in the correct order, your score will be (n2)(n3)(n-2)(n-3). Otherwise, it will be 0.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution:
Answer: ACGPRES
The smallest counterexamples are:
- Polya's conjecture: 906,150,257906,150,257
- Euler's sum of powers: 31,858,749,840,007,945,920,32131,858,749,840,007,945,920,321
- Cyclotomic polynomials: 105105
- Prime race: 23,338,590,79223,338,590,792
- Seventeen conjecture: 8,424,432,925,592,889,329,288,197,322,308,900,672,459,420,460,792,4338,424,432,925,592,889,329,288,197,322,308,900,672,459,420,460,792,433
- Goldbach's other conjecture: 57775777
- Average square: 4444

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.