Science:Math Exam Resources/Courses/MATH437/December 2011/Question 03

MATH437 December 2011

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Question 03
Show that there exist arbitrarily long sequences of consecutive positive integers such that no integer in the sequence is a power of a prime number.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Try to find a string of consecutive numbers where 2 divides the first number but 4 doesn't divide it, 3 divides the second number but 9 doesn't divide it, 5 divides the third number but 25 doesn't and so on through the primes.

Hint 2
The main idea uses the Chinese Remainder Theorem. Define a system of congruences by $\displaystyle x\equiv p_{i}-i+1\mod {p_{i}^{2}}$ for $\displaystyle i\in \{1,..,m-1\}$ where $\displaystyle p_{i}$ is the ith prime, m an integer and argue from here.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Let m be an arbitrary integer. The main idea uses the Chinese Remainder Theorem. Define a system of congruences by $\displaystyle x\equiv p_{i}-i+1\mod {p_{i}^{2}}$ for $\displaystyle i\in \{1,..,m-1\}$ where $\displaystyle p_{i}$ is the ith prime number. So $\displaystyle x\equiv 2-1+1\mod {4}$ $\displaystyle x\equiv 3-2+1\mod {9}$ $\displaystyle x\equiv 5-3+1\mod {25}$ and so on. Now look at the numbers $\displaystyle x,x+1,...,x+m-1$. Notice that for any $\displaystyle x+(i-1)$ for $\displaystyle i\in \{1,..,m-1\}$, we have $\displaystyle x+(i-1)\equiv p_{i}\mod {p_{i}^{2}}$ This means that $\displaystyle p_{i}\mid (x+i-1)$ however $\displaystyle p_{i}^{2}\nmid (x+i-1)$. Thus, no entry in this list can be a power of a prime (though some entries themselves might be prime if m is too small).

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