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4 votes
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Does $x^2+y$ represent infinitely many primes, where $|y|<|x|$?

All primes $p > 2$ can be represented this way. Let $x$ be the nearest integer to $\sqrt{p}$. So $|x -\sqrt{p}|< 1/2$, i.e. $x-1/2 < \sqrt{p} < x + 1/2$, and that says $x^2 - x + 1/4 < ...
Robert Israel's user avatar
3 votes

Does $x^2+y$ represent infinitely many primes, where $|y|<|x|$?

For each $x=n$, $y$ can take values in the range $[-n+1 \ , \ n-1]$. This for $x^2+y$ gives us the integers in the range $[n^2-n+1 \ , \ n^2+n-1]$. In total, the attainable integers are $$\bigcup_{n=1}...
Euclid's user avatar
  • 1,138
2 votes
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Homomorphism between $\mathbb{Z}[i]$ and $\mathbb{ Z/(p)}$

Yes, it is possible whenever $p\equiv_4 1$, because then there are integers $a,b$ such that $p=a^2+b^2$. Also, as you rightly say, this implies that there is an element like $i$ in $\Bbb Z/(p)$, as $$...
cansomeonehelpmeout's user avatar
2 votes

Homomorphism between $\mathbb{Z}[i]$ and $\mathbb{ Z/(p)}$

You could also use the fact that $(p-1)!\equiv -1 \mod p$. If $p=4n+1$ then $(p-1)!=(2n)!(2n+1)...(4n)=(2n)!(p-2n)...(p-1)$, thus $-1\equiv(p-1)!\equiv (2n)!(2n)!(-1)^{2n}\equiv (2n)!^2\mod p$. Thus, $...
dialegou's user avatar
  • 377

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