Pocklington's algorithm

Pocklington separates 3 different cases for p:

The first case, if ${\displaystyle p=4m+3}$, with ${\displaystyle m\in \mathbb {N} }$, the solution is ${\displaystyle x\equiv \pm a^{m+1}}$.

The second case, if ${\displaystyle p=8m+5}$, with ${\displaystyle m\in \mathbb {N} }$ and

1. ${\displaystyle a^{2m+1}\equiv 1}$, the solution is ${\displaystyle x\equiv \pm a^{m+1}}$.
2. ${\displaystyle a^{2m+1}\equiv -1}$, 2 is a (quadratic) non-residue so ${\displaystyle 4^{2m+1}\equiv -1}$. This means that ${\displaystyle (4a)^{2m+1}\equiv 1}$ so ${\displaystyle y\equiv \pm (4a)^{m+1}}$ is a solution of ${\displaystyle y^{2}\equiv 4a}$. Hence ${\displaystyle x\equiv \pm y/2}$ or, if y is odd, ${\displaystyle x\equiv \pm (p+y)/2}$.

The third case, if ${\displaystyle p=8m+1}$, put ${\displaystyle D\equiv -a}$, so the equation to solve becomes ${\displaystyle x^{2}+D\equiv 0}$. Now find by trial and error ${\displaystyle t_{1}}$ and ${\displaystyle u_{1}}$ so that ${\displaystyle N=t_{1}^{2}-Du_{1}^{2}}$ is a quadratic non-residue. Furthermore, let

${\displaystyle t_{n}={\frac {(t_{1}+u_{1}{\sqrt {D}})^{n}+(t_{1}-u_{1}{\sqrt {D}})^{n}}{2}},\qquad u_{n}={\frac {(t_{1}+u_{1}{\sqrt {D}})^{n}-(t_{1}-u_{1}{\sqrt {D}})^{n}}{2{\sqrt {D}}}}}$.

The following equalities now hold:

${\displaystyle t_{m+n}=t_{m}t_{n}+Du_{m}u_{n},\quad u_{m+n}=t_{m}u_{n}+t_{n}u_{m}\quad {\mbox{and}}\quad t_{n}^{2}-Du_{n}^{2}=N^{n}}$.

Supposing that p is of the form ${\displaystyle 4m+1}$ (which is true if p is of the form ${\displaystyle 8m+1}$), D is a quadratic residue and ${\displaystyle t_{p}\equiv t_{1}^{p}\equiv t_{1},\quad u_{p}\equiv u_{1}^{p}D^{(p-1)/2}\equiv u_{1}}$. Now the equations

${\displaystyle t_{1}\equiv t_{p-1}t_{1}+Du_{p-1}u_{1}\quad {\mbox{and}}\quad u_{1}\equiv t_{p-1}u_{1}+t_{1}u_{p-1}}$

give a solution ${\displaystyle t_{p-1}\equiv 1,\quad u_{p-1}\equiv 0}$.

Let ${\displaystyle p-1=2r}$. Then ${\displaystyle 0\equiv u_{p-1}\equiv 2t_{r}u_{r}}$. This means that either ${\displaystyle t_{r}}$ or ${\displaystyle u_{r}}$ is divisible by p. If it is ${\displaystyle u_{r}}$, put ${\displaystyle r=2s}$ and proceed similarly with ${\displaystyle 0\equiv 2t_{s}u_{s}}$. Not every ${\displaystyle u_{i}}$ is divisible by p, for ${\displaystyle u_{1}}$ is not. The case ${\displaystyle u_{m}\equiv 0}$ with m odd is impossible, because ${\displaystyle t_{m}^{2}-Du_{m}^{2}\equiv N^{m}}$ holds and this would mean that ${\displaystyle t_{m}^{2}}$ is congruent to a quadratic non-residue, which is a contradiction. So this loop stops when ${\displaystyle t_{l}\equiv 0}$ for a particular l. This gives ${\displaystyle -Du_{l}^{2}\equiv N^{l}}$, and because ${\displaystyle -D}$ is a quadratic residue, l must be even. Put ${\displaystyle l=2k}$. Then ${\displaystyle 0\equiv t_{l}\equiv t_{k}^{2}+Du_{k}^{2}}$. So the solution of ${\displaystyle x^{2}+D\equiv 0}$ is got by solving the linear congruence ${\displaystyle u_{k}x\equiv \pm t_{k}}$.

${\displaystyle x^{2}\equiv 43{\pmod {47}}.}$

This is the first case, according to the algorithm, ${\displaystyle x\equiv 43^{(47+1)/2}=43^{12}\equiv 2}$, but then ${\displaystyle x^{2}=2^{2}=4}$ not 43, so we should not apply the algorithm at all. The reason why the algorithm is not applicable is that a=43 is a quadratic non residue for p=47.

Solve the congruence

${\displaystyle x^{2}\equiv 18{\pmod {23}}.}$

The modulus is 23. This is ${\displaystyle 23=4\cdot 5+3}$, so ${\displaystyle m=5}$. The solution should be ${\displaystyle x\equiv \pm 18^{6}\equiv \pm 8{\pmod {23}}}$, which is indeed true: ${\displaystyle (\pm 8)^{2}\equiv 64\equiv 18{\pmod {23}}}$.

Solve the congruence

${\displaystyle x^{2}\equiv 10{\pmod {13}}.}$

The modulus is 13. This is ${\displaystyle 13=8\cdot 1+5}$, so ${\displaystyle m=1}$. Now verifying ${\displaystyle 10^{2m+1}\equiv 10^{3}\equiv -1{\pmod {13}}}$. So the solution is ${\displaystyle x\equiv \pm y/2\equiv \pm (4a)^{2}/2\equiv \pm 800\equiv \pm 7{\pmod {13}}}$. This is indeed true: ${\displaystyle (\pm 7)^{2}\equiv 49\equiv 10{\pmod {13}}}$.

Solve the congruence ${\displaystyle x^{2}\equiv 13{\pmod {17}}}$. For this, write ${\displaystyle x^{2}-13=0}$. First find a ${\displaystyle t_{1}}$ and ${\displaystyle u_{1}}$ such that ${\displaystyle t_{1}^{2}+13u_{1}^{2}}$ is a quadratic nonresidue. Take for example ${\displaystyle t_{1}=3,u_{1}=1}$. Now find ${\displaystyle t_{8}}$, ${\displaystyle u_{8}}$ by computing

${\displaystyle t_{2}=t_{1}t_{1}+13u_{1}u_{1}=9-13=-4\equiv 13{\pmod {17}},}$

${\displaystyle u_{2}=t_{1}u_{1}+t_{1}u_{1}=3+3\equiv 6{\pmod {17}}.}$

And similarly ${\displaystyle t_{4}=-299\equiv 7{\pmod {17}},u_{4}=156\equiv 3{\pmod {17}}}$ such that ${\displaystyle t_{8}=-68\equiv 0{\pmod {17}},u_{8}=42\equiv 8{\pmod {17}}.}$

Since ${\displaystyle t_{8}=0}$, the equation ${\displaystyle 0\equiv t_{4}^{2}+13u_{4}^{2}\equiv 7^{2}-13\cdot 3^{2}{\pmod {17}}}$ which leads to solving the equation ${\displaystyle 3x\equiv \pm 7{\pmod {17}}}$. This has solution ${\displaystyle x\equiv \pm 8{\pmod {17}}}$. Indeed, ${\displaystyle (\pm 8)^{2}=64\equiv 13{\pmod {17}}}$.