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Suppose we consider primes of the form $p = 1 \text { mod } 4$, so that $p = a^2 + b^2$, $a$ and $b$ being integers. Considering only the first quadrant, all $(a,b)$ pairs will be of the form (odd,even) or (even,odd).

Now if we consider $q = (a + n)^2 + (b + n)^2 = p + 2na + 2nb + 2n^2$, $n$ an integer, then $q = 1 \text { mod } 4$.

$1 \text { mod } 4$ as an arithmetic progression contains an infinite number of primes, and while every $q$ may not turn out to be a prime, one would still expect an infinite number of prime pairs $(p,q)$.

Has any work been done related to this? My original motivation was that this could give very close Gaussian primes, differing by a distance of just $n\sqrt2$, when $n$ is small.

Update: I came across a recent paper (Bounded gaps between Gaussian primes) where instead of $(a + n,b + n)$, the author considers $(a + n,b + m)$, with $(n,m)$ either $\text (odd,odd)$ or $\text (even,even)$, and then proves it for the case $(0,m)$, $m<=246$.

Suppose we consider primes of the form $p = 1 \text { mod } 4$, so that $p = a^2 + b^2$, $a$ and $b$ being integers. Considering only the first quadrant, all $(a,b)$ pairs will be of the form (odd,even) or (even,odd).

Now if we consider $q = (a + n)^2 + (b + n)^2 = p + 2na + 2nb + 2n^2$, $n$ an integer, then $q = 1 \text { mod } 4$.

$1 \text { mod } 4$ as an arithmetic progression contains an infinite number of primes, and while every $q$ may not turn out to be a prime, one would still expect an infinite number of prime pairs $(p,q)$.

Has any work been done related to this? My original motivation was that this could give very close Gaussian primes, differing by a distance of just $n\sqrt2$, when $n$ is small.

Update: I came across a recent paper (Bounded gaps between Gaussian primes) where instead of $(a + n,b + n)$, the author considers $(a + n,b + m)$, with $(n,m)$ either $\text (odd,odd)$ or $\text (even,even)$, and then proves it for the case $(0,m)$, $m>=246$.

Suppose we consider primes of the form $p = 1 \text { mod } 4$, so that $p = a^2 + b^2$, $a$ and $b$ being integers. Considering only the first quadrant, all $(a,b)$ pairs will be of the form (odd,even) or (even,odd).

Now if we consider $q = (a + n)^2 + (b + n)^2 = p + 2na + 2nb + 2n^2$, $n$ an integer, then $q = 1 \text { mod } 4$.

$1 \text { mod } 4$ as an arithmetic progression contains an infinite number of primes, and while every $q$ may not turn out to be a prime, one would still expect an infinite number of prime pairs $(p,q)$.

Has any work been done related to this? My original motivation was that this could give very close Gaussian primes, differing by a distance of just $n\sqrt2$, when $n$ is small.

Suppose we consider primes of the form $p = 1 \text { mod } 4$, so that $p = a^2 + b^2$, $a$ and $b$ being integers. Considering only the first quadrant, all $(a,b)$ pairs will be of the form (odd,even) or (even,odd).

Now if we consider $q = (a + n)^2 + (b + n)^2 = p + 2na + 2nb + 2n^2$, $n$ an integer, then $q = 1 \text { mod } 4$.

$1 \text { mod } 4$ as an arithmetic progression contains an infinite number of primes, and while every $q$ may not turn out to be a prime, one would still expect an infinite number of prime pairs $(p,q)$.

Has any work been done related to this? My original motivation was that this could give very close Gaussian primes, differing by a distance of just $n\sqrt2$, when $n$ is small.

Update: I came across a recent paper (Bounded gaps between Gaussian primes) where instead of $(a + n,b + n)$, the author considers $(a + n,b + m)$, with $(n,m)$ either $\text (odd,odd)$ or $\text (even,even)$, and then proves it for the case $(0,m)$, $m<=246$.

Suppose we consider primes of the form $p = 1 \text { mod } 4$, so that $p = a^2 + b^2$, $a$ and $b$ being integers. Considering only the first quadrant, apart from $(1,1)$ all other $(a,b)$ pairs will be of the form (odd,even) or (even,odd).

Now if we consider $q = (a + n)^2 + (b + n)^2 = p + 2na + 2nb + 2n^2$, $n$ an integer, then $q = 1 \text { mod } 4$.

$1 \text { mod } 4$ as an arithmetic progression contains an infinite number of primes, and while every $q$ may not turn out to be a prime, one would still expect an infinite number of prime pairs $(p,q)$.

Has any work been done related to this? My original motivation was that this could give very close Gaussian primes, differing by a distance of just $n\sqrt2$, when $n$ is small.

Suppose we consider primes of the form $p = 1 \text { mod } 4$, so that $p = a^2 + b^2$, $a$ and $b$ being integers. Considering only the first quadrant, all $(a,b)$ pairs will be of the form (odd,even) or (even,odd).

Now if we consider $q = (a + n)^2 + (b + n)^2 = p + 2na + 2nb + 2n^2$, $n$ an integer, then $q = 1 \text { mod } 4$.

$1 \text { mod } 4$ as an arithmetic progression contains an infinite number of primes, and while every $q$ may not turn out to be a prime, one would still expect an infinite number of prime pairs $(p,q)$.

Has any work been done related to this? My original motivation was that this could give very close Gaussian primes, differing by a distance of just $n\sqrt2$, when $n$ is small.