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I'm seeking to prove that the finite sum below is a composite number.

$$3^{2015} + 3^{2014}\cdot2 + 3^{2013}\cdot2^2 + ... + 3\cdot2^{2014} + 2^{2015}$$

The motivation for this is a question from Oxford's Math Admissions Test 2015 Q2 iv).

Is $3^{2015} − 2^{2015}$ a prime number? Explain your reasoning carefully.

Suggested answer: Using GP from (i), note that $$3^{2015} − 2^{2015} = (3^5 −2^5)(3^{2010} + 3^{2005}\cdot2^5 + 3^{2000}\cdot2^{10} + ... + 3^5\cdot2^{2005} + 2^{2010})$$

But neither factor is 1, so $3^{2015} − 2^{2015}$ is not prime. It's easy to see that both brackets are not $1$ (the first is $211$ and the second is a finite sum of constants greater than $1$). (I added an answer below, explaining why this method works.)

My original answer: $$3^{2015} − 2^{2015} = (3-2)(3^{2015} + 3^{2014}\cdot2 + 3^{2013}\cdot2^2 + ... + 3\cdot2^{2014} + 2^{2015})$$

The first bracket is 1, but I can't conclude whether the second is a prime number. I've considered the following options:

  1. Checking if the number is even. Every term in the expression except for $3^{2010}$ is even, as it is a product of a power of 3 (odd) and a power of 2 (even). $3^{2010}$ is odd. Hence, the sum is odd. We cannot immediately conclude that it is a composite number (with factor 2).
  2. Using prime divisibility tests. Summing the digits to check if they're divisible by $3$ and checking divisibility by prime numbers up to the square root of the number are unfeasible, as the number is just so huge.
  3. Factorizing further. There is no straightforward way to factor out a common factor from every term in the expression, as $3^{2010}$ and $2^{2010}$ do not have a common factor other than $1$.
  4. Using results derived in previous parts. It's of no use to know that $3$ is the only prime number that's one more than a square number and that $2$ is the only prime number that's one less than a cube number. The previous parts seem to be priming students to check whether the expressions in both brackets can be equal to 1 (i.e. show that the number is prime).

Is it possible to use my method to show that the sum in the second term is a composite number?

Bill Dubuque
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  • Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed. – Xander Henderson Aug 31 '24 at 20:26

3 Answers3

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Assuming that the intended question is "is the long sum prime":

We remark that $a\,|\,b$ implies that $(n^a-m^a)\,|\,(n^b-m^b)$. You used this remark in showing that $3^5-2^5$ divided $3^{2015}-2^{2015}$.

Now, it is easy (if a bit tedious) to see that $2015=5\times 13\times 31$. Thus each of $a_5=3^5-2^5,a_{13}=3^{13}-2^{13},a_{31}=3^{31}-2^{31}$ divides the original number.

We first claim that these are pairwise relatively prime. That follows from a standard result which is demonstrated here.

It follows that both $a_{13}$ and $a_{31}$ divide the long sum and, as they are relatively prime, the long sum has the product $a_{13}\times a_{31}$ as a factor, hence is not prime.

lulu
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  • Thank you so much! I'll try to make sense of the result you quoted when I work on Concrete Mathematics later on. –  Aug 28 '24 at 12:33
  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque Aug 28 '24 at 21:45
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    @BillDubuque I don't regard your proposed duplicate as a duplicate of this question. Granted that the phrasing of the question could have been better, but the OP understands that the underlying number is composite by remarking that $3^5-2^5$ was a factor. The question was to show that the quotient was also composite. I guess you could argue that the duplicate can easily be extended to consider the case where the exponent $n$ has multiple prime factors, but I think it is clear that this would have been a stretch for the OP. – lulu Aug 28 '24 at 21:50
  • That's not what the question says. In any case your version is also a dupe. – Bill Dubuque Aug 28 '24 at 21:52
  • @BillDubuque It is what the (revised) question says. And if you can find the dupe you mention, I'll delete my solution which, to be sure, was nothing more than routine arithmetic. – lulu Aug 28 '24 at 21:58
  • It is clear from the answer posted by the OP that your reading of the question is not what was intended. – Bill Dubuque Aug 28 '24 at 22:34
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    @BillDubuque Yes, the OP is very confused. No question about it. That is why I posted my solution. To recap: the OP had been faced with the problem of showing that $N=3^{2015}-2^{2015}$ was composite. They did this by pointing out that it was divisible by $3^5-2^5=211$, just as the duplicate would have wanted. They then decided that it was important to show that the quotient $\frac N{211}$ was also composite, it is not clear why but it is a fair question (I would say). Great confusion ensued. I am $90%$ convinced that I have posed and solved the question they intended. – lulu Aug 28 '24 at 22:47
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    I have voted to reopen the question. – lulu Aug 28 '24 at 22:48
  • That's unfortunate, since - as you should surely know - either way it is a dupe. Further, your answer does not apply to the question (as currently written) so it will likely confuse many readers. – Bill Dubuque Aug 28 '24 at 22:56
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    @BillDubuque Again, if you can find a dupe of the other question, I'll delete my solution. It certainly would not surprise me, but I did not find it. – lulu Aug 28 '24 at 22:59
  • You cannot delete it since it is acceptted, so I am not going to waste further time here. It is your responsibility to search for dupes before answering. – Bill Dubuque Aug 28 '24 at 23:00
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Each term in the second bracket is at least $1$, and there are at least two terms in the bracket. So, their sum cannot be $1$. Hence, the first bracket cannot be the full number on the right, and thus is a proper divisor.

EDIT If you factorise it as $$3^{2015} − 2^{2015} = (3-2)(3^{2015} + 3^{2014}\cdot2 + 3^{2013}\cdot2^2 + \cdots + 3\cdot2^{2014} + 2^{2015})$$ The part in the second bracket is precisely $3^{2015}-2^{2015}$. If you want to show that it is not a prime, you should rather just look at $3^{2015}-2^{2015}$, and figure out if it is a prime. But that is back to the same problem. There might be a way to do it that way, but the given answer is easier to do, and can be found by playing around a little bit more.

Hope this helps. :)

ultralegend5385
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We need to think outside the box, and consider the definition of prime numbers rigorously.

If $n$ is prime, $1$ and $n$ are the only factors of $n$. $1 \times n = n$. Consider $a \times m=n$

  • If $a=1$, $m=n$.
  • If $a \ne 1$, $m \ne n$. $\therefore n$ cannot be a prime number.

We don't need to let $a=3$, $b=2$, $n=2015$. We can let $a=3^5$, $b=2^5$, $n=403$.

$\because$ We cannot conclude that the second bracket is prime if the first bracket is 1 (as explained in the question), we need to make the first bracket $\ne1$. This makes the second bracket $\ne 3^{2015}-2^{2015}$.

$3^{2015}-2^{2015}$ thus has two factors that are not 1 (the first one is $211$, the second one is a finite sum of constants $\gt1$) and not $3^{2015}-2^{2015}$ (both need to be multiplied by a constant that is $\ne 1$ to get $3^{2015}-2^{2015}$.

$\therefore$ We can conclude that $3^{2015}-2^{2015}$ is not prime, without knowing whether the gnarly expression in the second bracket is prime.