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Motivated with this question I formulated following question :

Does $\lfloor \sqrt{p} \rfloor$ generate all natural numbers where $p$ is a prime number of the form $4k+3$ ?

I wrote Maxima implementation of this generator :

P:[];
N:[];
p:3;
while p<20000 do
(if (mod(p,4)=3) then P:append(P,[p]),
p:next_prime(p))$
for i from 1 thru length(P) do
(m:floor(sqrt(P[i])),
if (not member(m,N)) then N:append(N,[m]))$
N;

You can run this code here .

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Pedja
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1 Answers1

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$\lfloor \sqrt{p}\rfloor = n$ if and only if $$n\le \sqrt{p}<n+1$$ if and only if $$n^2\le p<(n+1)^2$$ OP demands a prime (of a particular type, no less) between each pair of consecutive squares. This is known as Legendre's conjecture and is open, even without the $4k+3$ restriction.

vadim123
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