# Primes + Inetvel + conjecture on primes

a) Can we establish a proof, there exists infinitely many primes of the form $n^2$ + 1. Why is the unit digit of such a prime p always 1 or 7? Is there any reasonable procedure or concept for the fact that the unit digit 7 occur essentially twice as often, when we identify the primes < 10000?

b) can we prove or disprove that, there exists an interval of the form [$n^2$, $(n+1)^2$] containing at least 1000 prime numbers.

c) We know that even integer > or = 4 can be written as sum of two primes and integers > 5 can be written as sum of three primes. Of course, those are conjectures. I am not asking the proof of those conjectures. I would like to know those statements are equivalent or not. If yes, how you will justify?

• It has not been established that there are infinitely many primes of the form $n^2+1$. What does "Inetvel" mean? – Gerry Myerson Nov 8 '12 at 11:40
• what's "inetvel" ? – Fattie Jul 10 '16 at 12:24

If every even $n\ge4$ can be written as a sum of two primes, then every integer $m\ge6$ can be written as either $2+r$ or $3+r$ where $r\ge4$ is even, hence $m$ can be written as a sum of three primes.

a) The unit digit is always 1 or seven because it is not possible to end up with a number of the form $n^2+1$ with a unit digit of 9 or 3, and 5, while possible, implies that the number is dividable by 5.

b) Plug in $n=1000000$ (It works as the interval).

c) They have nothing (intrinsically) to do with each other.

Answer to $(a)$: If the unit digit of $n$ is an odd number , then $n^2+1$ is divisible by $2.$ If the unit digit is $2$ or $8$ , then $n^2+1$ is divisible by $5.$ If the unit digit is $0$ ,then the last digit of $n^2+1$ is $1.$ If the unit digit is $4$ or $6$ , then the last digit of $n^2+1$ is $7.$

• Please find out how to type math on this site. The first principle is to surround mathematical notation with dollar signs, . No need to sign your name, either as it shows up beneath the post. – Kevin Arlin Nov 8 '12 at 11:36

a) Further to B Sahu's answer:

$$2=1^2+1$$; $$5=2^2+1$$. In the rest of this proof, let $$p$$ be a prime unequal to 2 or 5.

$$p$$ is coprime to 10, and so its unit digit can't be even or 5, and so is 1, 3, 7 or 9.

If $$p=n^2+1$$ then $$n^2$$ is even so $$n$$ is even. Working modulo 10 and checking the even possibilities for $$n$$:

• $$n=0$$ means $$p=0^2+1=1$$ so $$p$$'s units digit may be 1.
• $$n=2$$ or 8 means $$p=2^2+1=5$$ but the only prime case here is $$p=5$$, noted above.
• $$n=4$$ or 6 means $$p=4^2+1=7$$ so $$p$$'s units digit may be 7.

The fact that $$p=7$$ for $$2/5$$ of even $$n$$, but $$p=1$$ for only $$1/5$$ of even $$n$$ explains how come 7 shows up as a units digit twice as often as 1 does.

Not only that, but $$p=1\mod 10$$ only if $$n=0\mod 10$$, so $$100\mid n^2$$, so $$p=n^2+1=1\mod 100$$. Hence why, when $$p$$ does end in 1, $$p$$ ends in 01. And $$n=4$$ or $$6\mod10$$ means that $$n$$ is even, so $$4\mid n^2$$. As $$n^2=6\mod 10$$, that entails $$n^2=16\mod 20$$. Hence why, when $$p$$ ends in 7, $$p$$'s tens digit is odd.