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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?

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

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A partial answer:

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.

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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 $7.$

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  • $\begingroup$ 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. $\endgroup$ – Kevin Arlin Nov 8 '12 at 11:36
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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.

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