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

There exist infinitely many integers such that divides .

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 for infinitely many values of , so I should try to express 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 , which generates infinitely many natural numbers as outcomes of .

First idea that came to my mind, for factoring the denominator was in terms Gaussian Integers (numbers of form where are integers and ), 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:

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

Lemma:Let be a prime. The diophantine equation:is solvable if and only ifor

*Proof:**If the considered equation has a solution , then . Hence either or .*

*For , is a solution.*

*We show that there is a solution for each prime . Let us study the existence of an integral solution to the wee known diophantine equation: . *

*Observe that is odd [otherwise, , but no square leaves a remainder 3 when divided by 4]. Thus in the relation:*

*factors and have greatest common divisor 2, and consequently one of them is a doubled square (to be denoted by ) and the other one times a square (to be denoted by ).*

*The case and is impossible because it leads to a smaller solution of diophantine equation: . [method of infinite descent]*

*It follows that , and , therefore *

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Now we return back to main statement.

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Let , where and is a prime of form or .

Now let us consider another result:

Lemma:There are infinitely number of primes of the form .

**Proof : ***Suppose is an integer. Let be the smallest prime divisor of . *

*We define .*

*Since is odd, cannot be equal to . It is clear that is bigger than (otherwise ). *

*We know that has the form or . Since we have:*

*.*

*Using Fermat’s little Theorem we get . *

*If was of the form then is odd and therefore we obtain or which is a contradiction since is odd.*

*Hence, is of the form , now we can repeat the procedure replacing with and produce an infinite sequence of primes of the form .*

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From above lemma, we can say that, for infinitely many values of

I have expressed denominator in product form, what remains to prove is that the expression 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.**