Find the largest positive integer $n<100,$ such that there exists an arithmetic progression of positive integers $a_1,a_2,...,a_n$ with the following properties.

1) All numbers $a_2,a_3,...,a_{n-1}$ are powers of positive integers, that is numbers of the form $j^k,$ where $j\geq 1$ and $k\geq 2$ are integers.

2) The numbers $a_1$ and $a_{n}$ are not powers of positive integers.

The answer is 99.

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This solution seems to use the fact that 101 is a prime. Is this a necessary condition? What happens if the condition was changed to $n < 200$ , where $201$ is not a prime?

For n < 200, replace 101 and 199 with m and m+198 such that both are prime. E.g. 13, 211.

C Lim
- 7 years, 10 months ago

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For n < 200, replace 101 and 199 with m and m+198 such that both are prime. E.g. 13, 211.

Would that work? I thought part of the idea with 101 is the fact that 102,…,198 aren't multiples of it.

Peter Byers
- 7 years, 10 months ago

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Indeed. For this construction, the important aspect is that we have two primes $p_1 > p_2 > n$ and $p_1 - p_2 = n-1$ .

Clearly we must have $n$ odd. Is this a sufficient condition?

How about the case of $n$ even? I do not know if there exists a sequence of 98 terms which satisfy the conditions. My guess is yes, but I have no proof of the fact.

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How about the case of n even? I do not know if there exists a sequence of 98 terms which satisfy the conditions. My guess is yes, but I have no proof of the fact.

For a sequence of 98 terms, CL's same basic method could be used to find a k-value such that 227k, 229k, ... 419k, 421k is an appropriate sequence.

Peter Byers
- 7 years, 10 months ago

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@Peter Byers – Great modification! Now I'm kicking myself for not thinking of it.

Note that is it still an open conjecture if for any positive integer $n$ , there exist 2 primes $p$ and $q$ such that $p + 2n = q$ (i.e. the 2 primes that we need). You can see A020483 of OEIS for more details.

You're right of course. Sorry, I have to pick large enough primes, like 211, 409.

C Lim
- 7 years, 10 months ago

This looks like a good research topic. What subjects do you think will be needed for this? Number theory perhaps?

Ching Z
- 7 years, 10 months ago

What BASIC maths skills do i need to begin to understand this ? If I were to start learning this stuff from scratch, where do I need to begin ? I do have high school level mathematics but this is beyond anything I ever learned before .

Simon Cochrane
- 4 years, 4 months ago

What a hard problem! If I may ask a question, why must the progression be of the form $101k, 102k, ... 199k$ ? Why can't it be $k, 2k, 3k ...$ or $m, m+k, m+2k ...$ ?

Zhang Ning
- 7 years, 10 months ago

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He chose the sequence because it works. See the above discussion for what are the crucial aspects of the sequence that works.

Here is a simpler way of understanding the proof, which uses very similar idea. Starting with the sequence $101, 102, \ldots , 199$ (where once again the important criterion is that both 101 and 199 are prime), we want to inductively make each term (from the second) a power.

Base step: Starting with the second, we multiply the sequence by 102, which makes the second term $102^2$ , which is a power.

Induction step: Let the terms $a_2, a_3, \ldots a_k$ be powers. Let the LCM of the exponents be $M$ . Then, multiplying the entire sequence by $a_{k+1}^{M+1}$ makes all the terms $a_2, \ldots a_{k+1}$ a power term.

Hence, it is possible to make $a_2, a_3, \ldots a_{98}$ all powers. Now, we check that $a_1$ (respectively $a_{99}$ ) is not a power. This is because it is a multiple of 101 (resp 199), but not $101^2$ (resp $199^2$ ).

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Dang. I had the exact idea that if you have a small sequence that's all powers then multiplying it the next item by $a^{\frac{M+1}{k+1}}|]$ gives you a longer sequence of all powers. But when I tried it out on a numeric example I forgot to start with a sequence of powers and it didn't work so I moved onto other ideas.

Matt McNabb
- 7 years, 10 months ago

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Take the arithmetic progression $101k, 102k, \ldots, 199k$ , which has 99 terms. We wish to find a

ksuch that $102k, 102k, \ldots, 198k$ are powers of positive integers, while 101k and 199k aren't. To be specific, we letSdenote the set of primes < 198 and excluding 101, then pick k of the form:$k =\prod_{p \in S} p^{m_p} = 2^{m_2} \times 3^{m_3} \times 5^{m_5} \times\ldots$

for suitable non-negative integers $m_2, m_3, m_5, m_7, \ldots$ .

Pick any distinct primes $q_{102}, q_{103}, \ldots, q_{198}$ ; for every prime $p\in S$ solve the following congruences:

$\begin{aligned} m_p + \nu(102, p) &\equiv 0 \pmod {q_{102}}, \\ m_p + \nu(103, p) &\equiv 0 \pmod {q_{103}},\\ &\vdots \\ m_p + \nu(198, p) &\equiv 0 \pmod {q_{198}}. \end{aligned}$

Here, $\nu(n, p)$ is the largest

jsuch that $p^j | n$ . Note that since all prime factors of $102, \ldots, 198$ lie in S, we have:$102 = \prod_{p\in S} p^{\nu(102, p)}, \ 103 = \prod_{p\in S} p^{\nu(103, p)}, \ \ldots, \ 198 = \prod_{p\in S} p^{\nu(198, p)}.$

By Chinese Remainder Theorem, there's a solution for $m_p$ in the above congruences. Since $k =\prod_{p \in S} p^{m_p}$ , we see that:

$102k = \prod_{p\in S} p^{\nu(102, p)} p^{m_p} = \prod_{p\in S} p^{m_p + \nu(102, p)}$

is a $q_{102}$ -th power, and likewise for each of $i = 103, \ldots, 198$ , the multiple $ik$ is a $q_i$ -th power. Thus, we have a sequence which satisfies (1).

This sequence clearly satisfies (2) as well since 101 and 199 are prime and k is not a multiple of 101 or 199.