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I am still being taught maths at school, so all these fancy symbols like $\sum$ I have no idea about (although writing it out through mathjax, I think it possibly means the "sum" of something) but I believe to have made three potential discoveries of always finding a prime number:

$1. \qquad (2n + 1)^2 - 2 = p$

$2. \qquad n^3 - (n - 1)^3 = p$

$3. \qquad \text{The first number after $(2n)^2$ to have the number $7$ as its last digit is always prime.}$

I do not know if these are facts, but I do not know how to prove if any of these are true or false.

My Attempt:

$$(a + b)^2 = (a + b)(a + b) = a^2 + ab + ba + b^2 = a^2 + b^2 + 2ab \tag1$$ $$\implies (2n + 1)^2 = (2n)^2 + 1^2 + 2\times 2n\times 1 = 4n^2 + 4n + 1$$ $$\implies (2n + 1)^2 - 2 = 4n^2 + 4n + 1 - 2 = 4n^2 + 4n - 1= p$$ $$\implies 4n^2 + 4n - 1 - p = 4n^2 + 4n - (1 + p) = 0$$ $$\therefore n = \frac{-4 \pm \sqrt{16 - 16(1 + p)}}{8}$$ This is where I get stuck because any prime number must be greater than $1$ which implies that we will be square rooting a negative number in the numerator of the above fraction. Doing a little bit of research, the square root of a negative number is not "real" so how can there be real solutions for $n$ such that $(2n + 1)^2 - 2 = p?$ For example, $(2\times3 + 1)^2 - 2 = 47$ which is a prime number. I cannot see if I am doing something wrong here. Is this a contradiction of some sort?

$$(a - b)^3 = (a - b)(a - b)(a - b) = \cdots = a^3 - 3a^2b + 3ab^2 - b^3\tag2$$ $$\implies (n - 1)^3 = n^3 - 3\times n^2\times1 + 3\times n\times1^2 - 1^3 = n^3 - 3n^2 + 3n - 1$$ $$\implies n^3 - (n^3 - 3n^2 + 3n - 1) = n^3 - n^3 + 3n^2 - 3n + 1 = 3n^2 - 3n + 1 = p$$ $$\implies 3n^2 - 3n + (1 - p) = 0$$ $$\therefore n = \frac{3 \pm \sqrt {9 - 12(1 - p)}}{6}$$

Since $(1 - p) < 0$ then this implies that $12(1 - p) < 0$ and so we will be square rooting a positive number in the numerator of the above fraction because: $$\forall (x, y) \in \mathbb{Z}, \ x - (-y) = x + y > 0 : x = 9\land 12\mid y$$ But I am unsure that I completed this problem correctly because I used the same steps of working out compared to $(1)$, however I cannot find a problem here. For example: $$(\pm 3)^3 - 2^3 = 19 \qquad \text{ and } \qquad \pm 3 = \frac{3 \pm \sqrt {9 - 12(1 - 19)}}{6}$$

But as for $(3)$, I do not know where to begin. Could somebody please help me and show me step by step how to prove any one of these statements true or false? You do not have to prove all three statements true or false, but if you wish to do so, it would be much appreciated. I tried solving for $n$ via the quadratic formula, and completing the square just gave me a simplified fraction for $n$ and nothing more. If you are willing to use other mathematical techniques of solving for $n$, please be clear and try not to skip too many steps in the process. Thank you in advance.

Edit: Can I just say, this was my first question on the MSE and it said "be clear" and "be specific" and things like that, so I tried to make this question as understandable as possible, and for my first question, I got +10 which is AMAZING! Thank you so much all of you people for helping me with this. Your answers and explanations were very clear and helped me have a further understanding of these problems, and now I can use other methods of solving problems like these (or perhaps simply trial and error). I absolutely love this community even if some math questions are WAY beyond my years! Once again, thank you so much! :)

Mr Pie
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    Sorry, are you saying that $(2n+1)^2-2$ is always prime? But $11^2-2$ isn't. – lulu Sep 04 '17 at 11:25
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    A suggestion: in testing forms like this, you need to look at a lot of examples, not just the first two or three. There are a lot of small primes so if you only look at small numbers you get a distorted view of things. – lulu Sep 04 '17 at 11:27
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    Note: it's not too hard to prove that there is no non-constant polynomial $p(n)$ with integer coefficients that only takes prime values. Worth doing as an exercise. – lulu Sep 04 '17 at 11:28
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    @lulu Regarding the polynonial, $p: \mathbb N \to \mathbb N$, correct? – Andrew Tawfeek Sep 04 '17 at 11:39
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    @AndrewTawfeek Well, $\mathbb Z \to \mathbb Z$ will do. It's a divisibility claim, not a statement about positivity. – lulu Sep 04 '17 at 11:40
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    @AndrewTawfeek Historical trivia: Pretty sure this was first shown by Goldbach. In any case, it's old. And it's worth mentioning Euler's example $n^2+n+41$ which gives primes for $n\in {0,1,\cdots, 40}$. – lulu Sep 04 '17 at 11:43
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    @lulu I'll give the problem you offered a shot. And thats interesting! I don't know much about Goldbach admittingly -- I'm currently reading through Men of Mathematics and don't believe he's included. Also, thats definitely a formula worth remembing, thanks for bringing it up. :) – Andrew Tawfeek Sep 04 '17 at 11:49
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    For examples like the last one you can construct counterexamples as follows - since the final digit is controlled, take $n$ a multiple of $5$ so that $(2n)^2$ ends in a zero - then you know that "last digit $7$" means adding $7$. Then if $(2n)^2$ is a multiple of $7$ the sum will have a factor equal to $7$ and won't be prime. So $n$ a multiple of $35$ will provide counterexamples. This strategy of knowing some part of the result (like the last digit), identifying a divisor, and then controlling the bit you don't know about so that it shares the same divisor is useful to know. – Mark Bennet Sep 04 '17 at 11:53
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    any non-constant polynomial can't always be prime. –  Sep 04 '17 at 11:57
  • Thank you all for your comments. Lucky I'm not one of those guys that cannot take constructive criticism, but a mathematician must be prepared for feedback like this :) – Mr Pie Sep 04 '17 at 12:32
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    To prove any of them false it suffices to find a single counter-example. For 1. take n=5 (or n=8). For 2. take n=6 (or n=9). For 3 take n=6 (or n=9). There is no known simple formula for producing primes in spite of thousands of years of study. – DanielWainfleet Sep 04 '17 at 12:52
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    I think your solutions of the quadratic equations are wrong. $4n^2 + 4n - 1= p$ is satisfied by $n=1$ and $p=7$ , but $n = \frac{-4 \pm \sqrt{16 - 16(1 + p)}}{8}$ isn't – miracle173 Sep 04 '17 at 14:18

3 Answers3

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For the first proposition $n=5$ is a counterexample. $$(2\cdot 5 +1)^2 -2 = 7 \cdot 17$$

For the second proposition, $n=6$ is a counterexample. $$6^3-(6-1)^3=7\cdot 13$$

For the last proposition, $n=6$ is a counterexample.

$$(2\cdot 6)^2=144 \text { and } 147=3\cdot 7^2$$

Andrew Tawfeek
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    Oh, sorry about that. I must have overlooked that. I mean, $7\times17 = 119$ which looks like a prime number....yeah perhaps I need to test just a little bit more... – Mr Pie Sep 04 '17 at 11:26
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    @user477343 Please see my edit for the remaining propositions. – Andrew Tawfeek Sep 04 '17 at 11:30
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    Well, if you don't bound n>1, 2 can be contradicted with n=1. – Jesse P Francis Sep 04 '17 at 11:31
  • Ok thank you for those. – Mr Pie Sep 04 '17 at 11:32
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    @user477343 On the note of proving things like this though, I highly recommend reading up on a method of proof called induction. Good try though! – Andrew Tawfeek Sep 04 '17 at 11:34
  • @AndrewTawfeek thanks for that. I will look into it :) And also, it seems like every counter example is divisible by 7. – Mr Pie Sep 04 '17 at 12:35
  • @user477343 No divisible by $7$ is not necessary, for example for 1, $n=11$ gives $527=17\times 31$. For 2, $n=8$ gives $169=13^2$. For 3, $n=9$ gives $327=3\times 109$. – Especially Lime Sep 04 '17 at 14:41
  • @EspeciallyLime I know $7$ wasn't necessary, I was just lazy. Turned on my calculator and put $f(x)=\frac{(2x + 1)^2 - 2}{7}$ then scrolled through the table to spot integers. Factors of $7$ are easy to overlook so figured its worth a shot. For the final proposition though, it just happened to have a factor of $7$. – Andrew Tawfeek Sep 04 '17 at 19:17
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You can disprove such a statement by finding a counterexample.

  • For small values of $n$ you can do this manually. There are at least 2 counterexamples for $n\le 10$ for each of your claims.
  • Your claims are wrong for at least 50% of all $n\le100$. For larger $n$ these values increase. So if you select an arbitrary two digit number you have at least 50% chance to find a counterexample. You can use Wolfram Alpha to check if a number is a prime number.
  • To check a larger range of values you can write a program, e.g. in Python or Mathematica.
miracle173
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    Yes I have heard of Mathematica whilst browsing on this site, but thank you for this. How do they put it? "I give you a (+1)" :) – Mr Pie Sep 05 '17 at 01:00
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Disproving by simple counterexample is the right approach.

In your (1), the quadratic formula result has a sign error. It should be $\sqrt{16+16(1+p)}$, not $16-16(1+p)$.

Building solutions using the quadratic formula requires a lot of steps where there can be errors, so you should always test the result with a simple case, e.g. plug $n=1$ into (1), get $p=7$, and see if those values solve your quadratic.

Ѕᴀᴀᴅ
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L.R.
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