Kunerth's algorithm

Kunerth's algorithm is an algorithm for computing the modular square root of a given number.^[1][2] The algorithm does not require the factorization of the modulus, and relies on modular operations that is often easy when the given number is prime.

To find y from a given value

${\displaystyle B=y^{2}{\bmod {N}},}$

it takes the following steps:

1. find the modular square root of ${\displaystyle r\equiv {\sqrt {\pm N}}{\pmod {B}}}$. This step is quite easy, irrespectively of how big N when ${\displaystyle B}$ is a prime.
2. solve a quadratic equation associated with the modular square root of ${\displaystyle w^{2}=A\cdot z^{2}+B\cdot z+C}$. Most of Kunerth's examples in his original paper solve this equation by having C be a integer square and thus setting z to zero.

   Expand out the following equation to obtain the quadratic

   ${\displaystyle ((B\cdot z+r)^{2}\pm N)/B=w^{2}.}$

   One can always make sure that the quadratic can be solved by adjusting the modulus N in the above equation. Thus

   ${\displaystyle ((B\cdot z+r)^{2}+(B\cdot F+N))/B=w^{2}}$

   will ensure a quadratic of ${\displaystyle A\cdot z^{2}+D\cdot z+C+F}$.

   One can then adjust F to make sure that ${\displaystyle C+F}$ is a square. For large moduli, such as ${\displaystyle {\sqrt {67}}{\bmod {67F+RSA260}}}$, can have their square roots computed quickly via this method.

   The parameters of the polynomial expansion are quite flexible, in that ${\displaystyle ((67z+r)^{2}+X\cdot RSA260)/(67y)}$ can be done, for instance. It is quite easy to choose X and Y such that ${\displaystyle (r^{2}+X\cdot RSA260)/(67y)}$ is a square. The modular square root of ${\displaystyle {\sqrt {67y}}{\bmod {RSA260}}}$ can be taken this way.

3. Having solved the associated quadratic equation we now have the variables w and set v = r (if C in the quadratic is a natural square).
4. Solve for variables ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ the following equation:

   ${\displaystyle \alpha =w(v+w\beta ),}$

5. Obtain a value for X via factorization of the following polynomial:

   ${\displaystyle \alpha ^{2}\cdot x^{2}+(2\alpha \beta -N)x+(\beta ^{2}-(y^{2}{\bmod {N}}))}$

   obtaining an answer like

   ${\displaystyle (-37+9x)(1+25x)}$

6. Obtain the modular square root by the equation. Remember to set X such that the term above is zero. Thus X would be 37/9 or -1/25.

   ${\displaystyle y\equiv \alpha X+\beta {\pmod {N}}.}$

To obtain ${\displaystyle {\sqrt {41}}{\bmod {856}},}$ first obtain ${\displaystyle {\sqrt {-856}}\equiv 13{\pmod {41}}}$.

Then expand the polynomial:

${\displaystyle ((41z+13)^{2}+856)/41=w^{2}}$

into

${\displaystyle 25+26z+41z^{2}}$

Since, in this case the C term is a square, we take ${\displaystyle w=5}$ and compute ${\displaystyle v=13}$ (in general, ${\displaystyle v=41\cdot z+13}$).

Solve for ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ the following equation

${\displaystyle \alpha ==w(v+w\beta )}$

getting the solution ${\displaystyle \alpha =15}$ and ${\displaystyle \beta =-2}$. (There may be other pairs of solutions to this equation.)

Then factor the following polynomial:

${\displaystyle \alpha ^{2}x^{2}+(2\alpha \beta -856)x+(\beta ^{2}-41)}$

obtaining

${\displaystyle (-37+9x)(1+25x)}$

Then obtain the modular square root via

${\displaystyle 15\cdot (37\cdot 9^{-1})+(-2)\equiv 345{\pmod {856}}.}$

Verify that ${\displaystyle 345^{2}\equiv 41{\pmod {856}}.}$

In the case that ${\displaystyle {\sqrt {-856}}{\bmod {41}}}$ has no answer, then ${\displaystyle r\equiv {\sqrt {-856}}{\pmod {b\cdot 856+41}}}$ can be used instead.

• Methods of computing square roots

• Adolf Kunerth, "Sitzungsberichte. Academie Der Wissenschaften" vol 75, II, 1877, pp. 7–58
• Adolf Kunerth, "Sitzungsberichte. Academie Der Wissenschaften" vol 82, II, 1880, pp. 342–375