Are there 4 consecutive numbers that are each the sum of 2 squares? Some numbers can be expressed as the sum of two squares (ex. $10=3^2+1^2$) such as:
$$0,1,2,4,5,8,9,10,13,16,17,18,20,25,...$$
Other numbers are not the sum of any two squares of integers:
$$3,6,7,11,12,14,15,19,21,22,23,24,27,...$$
There are a lot of examples of $3$ consecutive numbers that are each the sum of 3 squares. The first is
$$0=0^2+0^2\quad\quad1=1^2+0^2 \quad\quad 2=1^2+1^2$$
The largest example I've found is
$$99952=444^2+896^2\quad 99953=568^2+823^2\quad 99954=327^2+945^2$$
This question seems to be related to the prime decomposition of consecutive numbers. It's pretty easy to use Fermat's theorem on sums of two squares and divisibility by $3$ to show that there can't be $6$ consecutive numbers of this type. But I can't manage to prove anything about $4$ or $5$ such consecutive numbers.
There were no examples of $4$ consecutive numbers of this type less than $100000$. Are there any?

No, there aren't
Every odd integer that is the sum of two squares, is congruent to $1$ modulo $4$ because every prime factor of the form $4k+3$ must occur in a power with even exponent.
