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Edited the formatting and added the quadratic-residues tag
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Dan Piponi
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Let $a(n)$ be the number of solutions of the equation $a^2+b^2\equiv -1 \pmod p_n$$a^2+b^2\equiv -1 \pmod {p_n}$, where $p_n$ is the n-th prime and $0\le a \le b \le \frac{p-1}2$$0\le a \le b \le \frac{p_n-1}2$. Is the sequence $a(1),a(2),a(3),\dots$ non-decreasing? Data for the first thousand values of the sequence supports this conjecture.

Here is an example for $n=5$:
The fifth prime is 11. The equation $a^2+b^2 \equiv -1 \pmod 11$$a^2+b^2 \equiv -1 \pmod {11}$ has just two solutions with the required conditions on $a$ and $b$, namely: $1^2+3^2=10$ and $4^2+4^2=32$.

Here are the first fifty values of $a(n)$: 0,1,1,1,2,2,3,3,3,4,4,5,6,6,6,7,8,8,9,9,10,10,11,12,13,13,13,14,14,15,16,17,18,18,19,19,20,21,21,22,23,23,24,25,25,25,27,28,29,29

Let $a(n)$ be the number of solutions of the equation $a^2+b^2\equiv -1 \pmod p_n$, where $p_n$ is the n-th prime and $0\le a \le b \le \frac{p-1}2$. Is the sequence $a(1),a(2),a(3),\dots$ non-decreasing? Data for the first thousand values of the sequence supports this conjecture.

Here is an example for $n=5$:
The fifth prime is 11. The equation $a^2+b^2 \equiv -1 \pmod 11$ has just two solutions with the required conditions on $a$ and $b$, namely: $1^2+3^2=10$ and $4^2+4^2=32$.

Here are the first fifty values of $a(n)$: 0,1,1,1,2,2,3,3,3,4,4,5,6,6,6,7,8,8,9,9,10,10,11,12,13,13,13,14,14,15,16,17,18,18,19,19,20,21,21,22,23,23,24,25,25,25,27,28,29,29

Let $a(n)$ be the number of solutions of the equation $a^2+b^2\equiv -1 \pmod {p_n}$, where $p_n$ is the n-th prime and $0\le a \le b \le \frac{p_n-1}2$. Is the sequence $a(1),a(2),a(3),\dots$ non-decreasing? Data for the first thousand values of the sequence supports this conjecture.

Here is an example for $n=5$:
The fifth prime is 11. The equation $a^2+b^2 \equiv -1 \pmod {11}$ has just two solutions with the required conditions on $a$ and $b$, namely: $1^2+3^2=10$ and $4^2+4^2=32$.

Here are the first fifty values of $a(n)$: 0,1,1,1,2,2,3,3,3,4,4,5,6,6,6,7,8,8,9,9,10,10,11,12,13,13,13,14,14,15,16,17,18,18,19,19,20,21,21,22,23,23,24,25,25,25,27,28,29,29

Edited the formatting and added the quadratic-residues tag
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Let a(n)$a(n)$ be the number of solutions of a^2+b^2 congruent tothe equation -1 modulo p$a^2+b^2\equiv -1 \pmod p_n$, where p=p(n)$p_n$ is the nthn-th prime and 0<=a<=b<=(p-1)/2$0\le a \le b \le \frac{p-1}2$. Is the sequence a(1),a(2),a(3),... always$a(1),a(2),a(3),\dots$ non-decreasing? Data for the first thousand values of the sequence supports this conjecture.

Here is an example for n=5$n=5$:
The fifth prime is 11. TheThe equation a^2+b^2 congruent to -1 (mod 11)$a^2+b^2 \equiv -1 \pmod 11$ has just two solutions with the required conditions on a$a$ and b$b$, namely: 1^2+3^2=10$1^2+3^2=10$ and 4^2+4^2=32$4^2+4^2=32$.

Here are the first fifty values of a(n)$a(n)$: 0,1,1,1,2,2,3,3,3,4,4,5,6,6,6,7,8,8,9,9,10,10,11,12,13,13,13,14,14,15,16,17,18,18,19,19,20,21,21,22,23,23,24,25,25,25,27,28,29,29

Let a(n) be the number of solutions of a^2+b^2 congruent to -1 modulo p, where p=p(n) is the nth prime and 0<=a<=b<=(p-1)/2. Is the sequence a(1),a(2),a(3),... always non-decreasing? Data for the first thousand values of the sequence supports this conjecture.

Here is an example for n=5:
The fifth prime is 11. The equation a^2+b^2 congruent to -1 (mod 11) has just two solutions with the required conditions on a and b, namely: 1^2+3^2=10 and 4^2+4^2=32.

Here are the first fifty values of a(n): 0,1,1,1,2,2,3,3,3,4,4,5,6,6,6,7,8,8,9,9,10,10,11,12,13,13,13,14,14,15,16,17,18,18,19,19,20,21,21,22,23,23,24,25,25,25,27,28,29,29

Let $a(n)$ be the number of solutions of the equation $a^2+b^2\equiv -1 \pmod p_n$, where $p_n$ is the n-th prime and $0\le a \le b \le \frac{p-1}2$. Is the sequence $a(1),a(2),a(3),\dots$ non-decreasing? Data for the first thousand values of the sequence supports this conjecture.

Here is an example for $n=5$:
The fifth prime is 11. The equation $a^2+b^2 \equiv -1 \pmod 11$ has just two solutions with the required conditions on $a$ and $b$, namely: $1^2+3^2=10$ and $4^2+4^2=32$.

Here are the first fifty values of $a(n)$: 0,1,1,1,2,2,3,3,3,4,4,5,6,6,6,7,8,8,9,9,10,10,11,12,13,13,13,14,14,15,16,17,18,18,19,19,20,21,21,22,23,23,24,25,25,25,27,28,29,29

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How does this sequence grow

Let a(n) be the number of solutions of a^2+b^2 congruent to -1 modulo p, where p=p(n) is the nth prime and 0<=a<=b<=(p-1)/2. Is the sequence a(1),a(2),a(3),... always non-decreasing? Data for the first thousand values of the sequence supports this conjecture.

Here is an example for n=5:
The fifth prime is 11. The equation a^2+b^2 congruent to -1 (mod 11) has just two solutions with the required conditions on a and b, namely: 1^2+3^2=10 and 4^2+4^2=32.

Here are the first fifty values of a(n): 0,1,1,1,2,2,3,3,3,4,4,5,6,6,6,7,8,8,9,9,10,10,11,12,13,13,13,14,14,15,16,17,18,18,19,19,20,21,21,22,23,23,24,25,25,25,27,28,29,29