# Prime Consequences

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Most of us are aware of the following consequence of Fundamental Theorem of Arithmetic:

There are infinitely many prime numbers.

The classic proof by Euclid is easy to follow. But I wanted to share the following two analytic equivalents (infinite series and infinite products) of the above purely arithmetical statement:

• $\displaystyle{\sum_{p}\frac{1}{p}}$   diverges.

For proof, refer to this discussion: https://math.stackexchange.com/q/361308/214604

• $\displaystyle{\sum_{n=1}^\infty \frac{1}{n^{s}} = \prod_p\left(1-\frac{1}{p^s}\right)^{-1}}$, where $s$ is any complex number with $\text{Re}(s)>1$.

The outline of proof,   when $s$ is a real number, has been discussed here: http://mathworld.wolfram.com/EulerProduct.html

# Popular-Lonely primes understood

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While reading standup mathematician Matt Parker‘s book Things to Make and do in Fourth Dimension, I found answer (on pp. 146) to the question I raised 7 months ago.

When the grid happens to be a multiple of 6 wide, suddenly all primes snap into dead-straight lines. All primes (except 2 and 3) are one more or less than a multiple of 6. (© Matt Parker, 2014)

He also proves the following surprising theorem:

The square of every prime number greater than 3 is one more than a multiple of 24.

Let $p$ be an odd prime not equal to 3. Now we subtract one from the square of this prime number. Therefore, we wish to prove that $p^2-1=(p-1)(p+1)$ is a multiple of 24.

Note that, $p^2-1$ is a product of two even numbers. In particular, one of these two even numbers must be a multiple of 4, as they are consecutive even numbers and every other even number is divisible by 4. Hence we conclude that $p^2-1$ is divisible by 8.

Observe that exactly one of three consecutive numbers, $p-1,p,p+1$ must be divisible by 3. Since $p$ is an odd prime different from 3, one of $p-1$ or $p+1$ must be divisible by 3. Hence we conclude that $p^2-1$ is divisible by 3.

Combining both the conclusions made above, we complete proof of our statement (since 2 and 3 are coprime).

Edit[19 April 2017]: Today I discovered that this theorem is exercise 68 in “The USSR Olympiad Problem Book“.

# Primes: popular and lonely

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ulam While doodling in class, I made a 10 x 10 grid and filled it with numbers from 1 to 100. The motivations behind 10 x 10 grid was human bias towards the number 10.

Then inspired by Ulam Spiral, I started creating paths (allowing diagonal, horizontal and vertical moves) starting from the smallest number. Following paths emerged:

• 2→ 3 →13 → 23
• 2 → 11
• 7 → 17
• 19 → 29
• 31 → 41
• 37 → 47
• 43 → 53
• 61 → 71
• 73 → 83
• 79 → 89

So, longest path is of length 4 and others are of length 2.

The number 2 is special one here, since it leads to two paths. I will call such primes, with more than one paths, popular primes.

Now, 5, 59, 67 and 97 don’t have any prime number neighbour. I will call such primes, with no neighbour, lonely primes.

I hope to create other $b \times b$ grids filled with 1 to $b^2$ natural numbers written in base $b$. Then will try to identify such lonely and popular primes.

# Diophantine Equation & Primes

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I stumbled upon following statement on some webpage (don’t have link):

There exist infinitely many integers $n$ such that $n^2 +1$ divides $n!$.

I tried proving it, though not able to prove it but here are my insights:

• Proof by contradiction (the way we show infinite primes) doesn’t work here.
• I need to prove that $\frac{n!}{n^2+1} = k \in \mathbb{N}$ for infinitely many values of $n$, so I should try to express $n^2+1$ in product form (since numerator is already in product form). After writing denominator in product form,  find a condition satisfied by infinitely many values of $n$, which generates infinitely many natural numbers as outcomes of $\frac{n!}{n^2+1}$.

First idea that came to my mind, for factoring the denominator was in terms Gaussian Integers (numbers of form $a+ib$ where $a,b$ are integers and $i = \sqrt{-1}$), but I am not able to proceed in this direction further.

Second idea to generate factors of denominator was to analyze Diophantine equation (i.e. we are concerned with only integer solutions of given equation), of form: $x^2 + 1 = Dy^2$

Here is a theorem from Diophantine equation, which can help me generate factors:

Lemma: Let $p$ be a prime. The diophantine equation: $x^2 - py^2 = -1$ is solvable if and only if $p=2$ or $p\equiv 1 \pmod 4$

Proof: If the considered equation has a solution $(x,y)$, then $p| x^2+1$. Hence either  $p=2$ or  $p \equiv 1 \pmod 4$.

For $p=2$$x=y=1$ is a solution.

We show that there is a solution for each prime $p=4t+1$. Let us study the existence of an integral solution $(x_0,y_0)$ to the wee known diophantine equation: $x_0^2 - py_0^2 = 1$.

Observe that $x_0$ is odd [otherwise, $y_0^2 \equiv py_0^2 \equiv 3 \pmod 4$, but no square leaves a remainder 3 when divided by 4]. Thus in the relation:

$x_0^2 - 1 = (x_0-1)(x_0+1) = py_0^2$

factors $x_0+1$ and $x_0-1$ have greatest common divisor 2, and consequently one of them is a doubled square (to be denoted by $2x^2$) and the other one $2p$ times a square (to be denoted by $2py^2$).

The case $x_0+1 = 2x^2$ and $x_0-1 = 2py^2$ is impossible because it leads to a smaller solution of diophantine equation: $x^2-py^2=1$. [method of infinite descent]

It follows that $x_0-1=2x^2$ , and $x_0+1 = 2py^2$, therefore $x^2-py^2 = -1$

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Now we return back to main statement.

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Let $n^2+1 = pm^2$, where $m \in \mathbb{N}$ and $p$ is a prime of  form $p=2$ or $p \equiv1 \pmod 4$.

Now let us consider another result:

Lemma: There are infinitely number of primes of the form $4n+1$.

ProofSuppose $n>1$ is an integer. Let $p$ be the smallest prime divisor of $N$.

We define $N=(n!)^2 +1$.

Since $N$ is odd, $p$ cannot be equal to $2$. It is clear that $p$ is bigger than $n$ (otherwise $p \mid 1$ ).

We know that $p$ has the form $4k+1$ or $4k+3$. Since $p\mid N$ we have:

$(n!)^2 \equiv -1 \pmod p \$

$\Rightarrow (n!)^{p-1} \equiv (-1)^{ \frac{p-1}{2} } \pmod p$.

Using Fermat’s little Theorem we get $(-1)^{ \frac{p-1}{2} } \equiv 1 \pmod p$.

If $p$ was of the form $4k+3$ then $\frac{p-1}{2} =2k+1$ is odd and therefore we obtain $-1 \equiv 1 \pmod p$ or $p \mid 2$ which is a contradiction since $p$ is odd.

Hence, $p$ is of the form $4k+1$, now we can repeat the procedure replacing $n$ with $p$ and produce an infinite sequence of primes of the form $4k+1$.

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From above lemma, we can say that, $n^2+1 = pm^2$ for infinitely many values of $p$

I have expressed denominator in product form, what remains to prove is that the expression $\frac{n!}{pm^2}$  simplifies to give a natural number as answer.

Now this appears to me as a dead end.

If you discover/know the proof of starting statement please give its outline in comments.

But, the motive of this post is to show the beautiful flowers I observed while searching for a path which leads to given statement.