Menu
Home
People
Places
Arts
History
Plants & Animals
Science
Life & Culture
Technology
Reference.org
Cipolla's algorithm
open-in-new
<h2 id="algorithm">Algorithm</h2> <p>Inputs: </p> <ul><li> p {\displaystyle p} , an odd prime,</li> <li> n ∈ F p {\displaystyle n\in \mathbf {F} _{p}} , which is a square.</li></ul> <p>Outputs: </p> <ul><li> x ∈ F p {\displaystyle x\in \mathbf {F} _{p}} , satisfying x 2 = n . {\displaystyle x^{2}=n.} </li></ul> <p>Step 1 is to find an a ∈ F p {\displaystyle a\in \mathbf {F} _{p}} such that a 2 − n {\displaystyle a^{2}-n} is not a square. There is no known deterministic algorithm for finding such an a {\displaystyle a} , but the following <a href="/facts/Trial_and_error/QdxUKRld">trial and error</a> method can be used. Simply pick an a {\displaystyle a} and by computing the <a href="/facts/Legendre_symbol/MwHrJ7as">Legendre symbol</a> ( a 2 − n p ) {\displaystyle \left({\frac {a^{2}-n}{p}}\right)} one can see whether a {\displaystyle a} satisfies the condition. The chance that a random a {\displaystyle a} will satisfy is ( p − 1 ) / 2 p {\displaystyle (p-1)/2p} . With p {\displaystyle p} large enough this is about 1 / 2 {\displaystyle 1/2} .<a class="footnote-ref" id="fnref:2" href="#fn:2"><sup>2</sup></a> Therefore, the expected number of trials before finding a suitable a {\displaystyle a} is about 2. </p><p>Step 2 is to compute <i>x</i> by computing x = ( a + a 2 − n ) ( p + 1 ) / 2 {\displaystyle x=\left(a+{\sqrt {a^{2}-n}}\right)^{(p+1)/2}} within the <a href="/facts/Field_extension/L7yIDtyK">field extension</a> F p 2 = F p ( a 2 − n ) {\displaystyle \mathbf {F} _{p^{2}}=\mathbf {F} _{p}({\sqrt {a^{2}-n}})} . This <i>x</i> will be the one satisfying x 2 = n . {\displaystyle x^{2}=n.} </p><p>If x 2 = n {\displaystyle x^{2}=n} , then ( − x ) 2 = n {\displaystyle (-x)^{2}=n} also holds. And since <i>p</i> is odd, x ≠ − x {\displaystyle x\neq -x} . So whenever a solution <i>x</i> is found, there's always a second solution, <i>-x</i>. </p> <h2 id="example">Example</h2> <p>(Note: All elements before step two are considered as an element of F 13 {\displaystyle \mathbf {F} _{13}} and all elements in step two are considered as elements of F 13 2 {\displaystyle \mathbf {F} _{13^{2}}} .) </p><p>Find all <i>x</i> such that x 2 = 10. {\displaystyle x^{2}=10.} </p><p>Before applying the algorithm, it must be checked that 10 {\displaystyle 10} is indeed a square in F 13 {\displaystyle \mathbf {F} _{13}} . Therefore, the Legendre symbol ( 10 | 13 ) {\displaystyle (10|13)} has to be equal to 1. This can be computed using <a href="/facts/Euler%2527s_criterion/JkbpqqB1">Euler's criterion</a>: ( 10 | 13 ) ≡ 10 6 ≡ 1 ( mod 13 ) . {\textstyle (10|13)\equiv 10^{6}\equiv 1{\pmod {13}}.} This confirms 10 being a square and hence the algorithm can be applied. </p> <ul><li>Step 1: Find an <i>a</i> such that a 2 − n {\displaystyle a^{2}-n} is not a square. As stated, this has to be done by trial and error. Choose a = 2 {\displaystyle a=2} . Then a 2 − n {\displaystyle a^{2}-n} becomes 7. The Legendre symbol ( 7 | 13 ) {\displaystyle (7|13)} has to be −1. Again this can be computed using Euler's criterion: 7 6 = 343 2 ≡ 5 2 ≡ 25 ≡ − 1 ( mod 13 ) . {\textstyle 7^{6}=343^{2}\equiv 5^{2}\equiv 25\equiv -1{\pmod {13}}.} So a = 2 {\displaystyle a=2} is a suitable choice for <i>a</i>.</li> <li>Step 2: Compute x = ( a + a 2 − n ) ( p + 1 ) / 2 = ( 2 + − 6 ) 7 {\displaystyle x=\left(a+{\sqrt {a^{2}-n}}\right)^{(p+1)/2}=\left(2+{\sqrt {-6}}\right)^{7}} in F 13 ( − 6 ) {\displaystyle \mathbf {F} _{13}({\sqrt {-6}})} :</li></ul> ( 2 + − 6 ) 2 = 4 + 4 − 6 − 6 = − 2 + 4 − 6 {\displaystyle \left(2+{\sqrt {-6}}\right)^{2}=4+4{\sqrt {-6}}-6=-2+4{\sqrt {-6}}} ( 2 + − 6 ) 4 = ( − 2 + 4 − 6 ) 2 = − 1 − 3 − 6 {\displaystyle \left(2+{\sqrt {-6}}\right)^{4}=\left(-2+4{\sqrt {-6}}\right)^{2}=-1-3{\sqrt {-6}}} ( 2 + − 6 ) 6 = ( − 2 + 4 − 6 ) ( − 1 − 3 − 6 ) = 9 + 2 − 6 {\displaystyle \left(2+{\sqrt {-6}}\right)^{6}=\left(-2+4{\sqrt {-6}}\right)\left(-1-3{\sqrt {-6}}\right)=9+2{\sqrt {-6}}} ( 2 + − 6 ) 7 = ( 9 + 2 − 6 ) ( 2 + − 6 ) = 6 {\displaystyle \left(2+{\sqrt {-6}}\right)^{7}=\left(9+2{\sqrt {-6}}\right)\left(2+{\sqrt {-6}}\right)=6} <p>So x = 6 {\displaystyle x=6} is a solution, as well as x = − 6 {\displaystyle x=-6} . Indeed, 6 2 ≡ 10 ( mod 13 ) . {\textstyle 6^{2}\equiv 10{\pmod {13}}.} </p> <h2 id="proof">Proof</h2> <p>The first part of the proof is to verify that F p 2 = F p ( a 2 − n ) = { x + y a 2 − n : x , y ∈ F p } {\displaystyle \mathbf {F} _{p^{2}}=\mathbf {F} _{p}({\sqrt {a^{2}-n}})=\{x+y{\sqrt {a^{2}-n}}:x,y\in \mathbf {F} _{p}\}} is indeed a field. For the sake of notation simplicity, ω {\displaystyle \omega } is defined as a 2 − n {\displaystyle {\sqrt {a^{2}-n}}} . Of course, a 2 − n {\displaystyle a^{2}-n} is a quadratic non-residue, so there is no <a href="/facts/Square_root/AfrzfBdQ">square root</a> in F p {\displaystyle \mathbf {F} _{p}} . This ω {\displaystyle \omega } can roughly be seen as analogous to the complex number <a href="/facts/Imaginary_unit/1SCllQlU">i</a>. The field arithmetic is quite obvious. <a href="/facts/Addition/apFWJBFB">Addition</a> is defined as </p> ( x 1 + y 1 ω ) + ( x 2 + y 2 ω ) = ( x 1 + x 2 ) + ( y 1 + y 2 ) ω {\displaystyle \left(x_{1}+y_{1}\omega \right)+\left(x_{2}+y_{2}\omega \right)=\left(x_{1}+x_{2}\right)+\left(y_{1}+y_{2}\right)\omega } . <p><a href="/facts/Multiplication/VtcUjpNv">Multiplication</a> is also defined as usual. With keeping in mind that ω 2 = a 2 − n {\displaystyle \omega ^{2}=a^{2}-n} , it becomes </p> ( x 1 + y 1 ω ) ( x 2 + y 2 ω ) = x 1 x 2 + x 1 y 2 ω + y 1 x 2 ω + y 1 y 2 ω 2 = ( x 1 x 2 + y 1 y 2 ( a 2 − n ) ) + ( x 1 y 2 + y 1 x 2 ) ω {\displaystyle \left(x_{1}+y_{1}\omega \right)\left(x_{2}+y_{2}\omega \right)=x_{1}x_{2}+x_{1}y_{2}\omega +y_{1}x_{2}\omega +y_{1}y_{2}\omega ^{2}=\left(x_{1}x_{2}+y_{1}y_{2}\left(a^{2}-n\right)\right)+\left(x_{1}y_{2}+y_{1}x_{2}\right)\omega } . <p>Now the field properties have to be checked. The properties of closure under addition and multiplication, <a href="/facts/Associativity/EQX8lV6r">associativity</a>, <a href="/facts/Commutativity/WaYxJLxd">commutativity</a> and <a href="/facts/Distributivity/0LNBrVJl">distributivity</a> are easily seen. This is because in this case the field F p 2 {\displaystyle \mathbf {F} _{p^{2}}} is somewhat resembles the field of <a href="/facts/Complex_number/2w7mqI7e">complex numbers</a> (with ω {\displaystyle \omega } being the analogon of <i>i</i>). The additive <a href="/facts/Identity_element/9nss3xHY">identity</a> is 0 {\displaystyle 0} , or more formally 0 + 0 ω {\displaystyle 0+0\omega } : Let α ∈ F p 2 {\displaystyle \alpha \in \mathbf {F} _{p^{2}}} , then </p> α + 0 = ( x + y ω ) + ( 0 + 0 ω ) = ( x + 0 ) + ( y + 0 ) ω = x + y ω = α {\displaystyle \alpha +0=(x+y\omega )+(0+0\omega )=(x+0)+(y+0)\omega =x+y\omega =\alpha } . <p>The multiplicative identity is 1 {\displaystyle 1} , or more formally 1 + 0 ω {\displaystyle 1+0\omega } : </p> α ⋅ 1 = ( x + y ω ) ( 1 + 0 ω ) = ( x ⋅ 1 + 0 ⋅ y ( a 2 − n ) ) + ( x ⋅ 0 + 1 ⋅ y ) ω = x + y ω = α {\displaystyle \alpha \cdot 1=(x+y\omega )(1+0\omega )=\left(x\cdot 1+0\cdot y\left(a^{2}-n\right)\right)+(x\cdot 0+1\cdot y)\omega =x+y\omega =\alpha } . <p>The only thing left for F p 2 {\displaystyle \mathbf {F} _{p^{2}}} being a field is the existence of additive and multiplicative <a href="/facts/Inverse_element/CwMygHSj">inverses</a>. It is easily seen that the additive inverse of x + y ω {\displaystyle x+y\omega } is − x − y ω {\displaystyle -x-y\omega } , which is an element of F p 2 {\displaystyle \mathbf {F} _{p^{2}}} , because − x , − y ∈ F p {\displaystyle -x,-y\in \mathbf {F} _{p}} . In fact, those are the additive inverse elements of <i>x</i> and <i>y</i>. For showing that every non-zero element α {\displaystyle \alpha } has a multiplicative inverse, write down α = x 1 + y 1 ω {\displaystyle \alpha =x_{1}+y_{1}\omega } and α − 1 = x 2 + y 2 ω {\displaystyle \alpha ^{-1}=x_{2}+y_{2}\omega } . In other words, </p> ( x 1 + y 1 ω ) ( x 2 + y 2 ω ) = ( x 1 x 2 + y 1 y 2 ( a 2 − n ) ) + ( x 1 y 2 + y 1 x 2 ) ω = 1 {\displaystyle (x_{1}+y_{1}\omega )(x_{2}+y_{2}\omega )=\left(x_{1}x_{2}+y_{1}y_{2}\left(a^{2}-n\right)\right)+\left(x_{1}y_{2}+y_{1}x_{2}\right)\omega =1} . <p>So the two equalities x 1 x 2 + y 1 y 2 ( a 2 − n ) = 1 {\displaystyle x_{1}x_{2}+y_{1}y_{2}(a^{2}-n)=1} and x 1 y 2 + y 1 x 2 = 0 {\displaystyle x_{1}y_{2}+y_{1}x_{2}=0} must hold. Working out the details gives expressions for x 2 {\displaystyle x_{2}} and y 2 {\displaystyle y_{2}} , namely </p> x 2 = − y 1 − 1 x 1 ( y 1 ( a 2 − n ) − x 1 2 y 1 − 1 ) − 1 {\displaystyle x_{2}=-y_{1}^{-1}x_{1}\left(y_{1}\left(a^{2}-n\right)-x_{1}^{2}y_{1}^{-1}\right)^{-1}} , y 2 = ( y 1 ( a 2 − n ) − x 1 2 y 1 − 1 ) − 1 {\displaystyle y_{2}=\left(y_{1}\left(a^{2}-n\right)-x_{1}^{2}y_{1}^{-1}\right)^{-1}} . <p>The inverse elements which are shown in the expressions of x 2 {\displaystyle x_{2}} and y 2 {\displaystyle y_{2}} do exist, because these are all elements of F p {\displaystyle \mathbf {F} _{p}} . This completes the first part of the proof, showing that F p 2 {\displaystyle \mathbf {F} _{p^{2}}} is a field. </p><p>The second and middle part of the proof is showing that for every element x + y ω ∈ F p 2 : ( x + y ω ) p = x − y ω {\displaystyle x+y\omega \in \mathbf {F} _{p^{2}}:(x+y\omega )^{p}=x-y\omega } . By definition, ω 2 = a 2 − n {\displaystyle \omega ^{2}=a^{2}-n} is not a square in F p {\displaystyle \mathbf {F} _{p}} . Euler's criterion then says that </p> ω p − 1 = ( ω 2 ) p − 1 2 = − 1 {\displaystyle \omega ^{p-1}=\left(\omega ^{2}\right)^{\frac {p-1}{2}}=-1} . <p>Thus ω p = − ω {\displaystyle \omega ^{p}=-\omega } . This, together with <a href="/facts/Fermat%2527s_little_theorem/qIjSIm2s">Fermat's little theorem</a> (which says that x p = x {\displaystyle x^{p}=x} for all x ∈ F p {\displaystyle x\in \mathbf {F} _{p}} ) and the knowledge that in fields of <a href="/facts/Characteristic_(algebra)/D2VzcQaG">characteristic</a> <i>p</i> the equation ( a + b ) p = a p + b p {\displaystyle \left(a+b\right)^{p}=a^{p}+b^{p}} holds, a relationship sometimes called the <a href="/facts/Freshman%2527s_dream/uph8o5RB">Freshman's dream</a>, shows the desired result </p> ( x + y ω ) p = x p + y p ω p = x − y ω {\displaystyle (x+y\omega )^{p}=x^{p}+y^{p}\omega ^{p}=x-y\omega } . <p>The third and last part of the proof is to show that if x 0 = ( a + ω ) p + 1 2 ∈ F p 2 {\displaystyle x_{0}=\left(a+\omega \right)^{\frac {p+1}{2}}\in \mathbf {F} _{p^{2}}} , then x 0 2 = n ∈ F p {\displaystyle x_{0}^{2}=n\in \mathbf {F} _{p}} . Compute </p> x 0 2 = ( a + ω ) p + 1 = ( a + ω ) ( a + ω ) p = ( a + ω ) ( a − ω ) = a 2 − ω 2 = a 2 − ( a 2 − n ) = n {\displaystyle x_{0}^{2}=\left(a+\omega \right)^{p+1}=(a+\omega )(a+\omega )^{p}=(a+\omega )(a-\omega )=a^{2}-\omega ^{2}=a^{2}-\left(a^{2}-n\right)=n} . <p>Note that this computation took place in F p 2 {\displaystyle \mathbf {F} _{p^{2}}} , so this x 0 ∈ F p 2 {\displaystyle x_{0}\in \mathbf {F} _{p^{2}}} . But with <a href="/facts/Lagrange%2527s_theorem_(number_theory)/kCXB0FJH">Lagrange's theorem</a>, stating that a non-zero <a href="/facts/Integer_polynomial/Lzak8VVx">polynomial</a> of degree <i>n</i> has at most <i>n</i> roots in any field <i>K</i>, and the knowledge that x 2 − n {\displaystyle x^{2}-n} has 2 roots in F p {\displaystyle \mathbf {F} _{p}} , these roots must be all of the roots in F p 2 {\displaystyle \mathbf {F} _{p^{2}}} . It was just shown that x 0 {\displaystyle x_{0}} and − x 0 {\displaystyle -x_{0}} are roots of x 2 − n {\displaystyle x^{2}-n} in F p 2 {\displaystyle \mathbf {F} _{p^{2}}} , so it must be that x 0 , − x 0 ∈ F p {\displaystyle x_{0},-x_{0}\in \mathbf {F} _{p}} .<a class="footnote-ref" id="fnref:3" href="#fn:3"><sup>3</sup></a> </p> <h2 id="speed">Speed</h2> <p>After finding a suitable <i>a</i>, the number of operations required for the algorithm is 4 m + 2 k − 4 {\displaystyle 4m+2k-4} multiplications, 4 m − 2 {\displaystyle 4m-2} sums, where <i>m</i> is the number of <a href="/facts/Numerical_digit/JJdLBoGT">digits</a> in the <a href="/facts/Binary_numeral_system/a6JDSRPs">binary representation</a> of <i>p</i> and <i>k</i> is the number of ones in this representation. To find <i>a</i> by trial and error, the expected number of computations of the Legendre symbol is 2. But one can be lucky with the first try and one may need more than 2 tries. In the field F p 2 {\displaystyle \mathbf {F} _{p^{2}}} , the following two equalities hold </p> ( x + y ω ) 2 = ( x 2 + y 2 ω 2 ) + ( ( x + y ) 2 − x 2 − y 2 ) ω , {\displaystyle (x+y\omega )^{2}=\left(x^{2}+y^{2}\omega ^{2}\right)+\left(\left(x+y\right)^{2}-x^{2}-y^{2}\right)\omega ,} <p>where ω 2 = a 2 − n {\displaystyle \omega ^{2}=a^{2}-n} is known in advance. This computation needs 4 multiplications and 4 sums. </p> ( x + y ω ) 2 ( a + ω ) = ( a d 2 − b ( x + d ) ) + ( d 2 − b y ) ω , {\displaystyle \left(x+y\omega \right)^{2}\left(a+\omega \right)=\left(ad^{2}-b\left(x+d\right)\right)+\left(d^{2}-by\right)\omega ,} <p>where d = ( x + y a ) {\displaystyle d=(x+ya)} and b = n y {\displaystyle b=ny} . This operation needs 6 multiplications and 4 sums. </p><p>Assuming that p ≡ 1 ( mod 4 ) , {\displaystyle p\equiv 1{\pmod {4}},} (in the case p ≡ 3 ( mod 4 ) {\displaystyle p\equiv 3{\pmod {4}}} , the direct computation x ≡ ± n p + 1 4 {\displaystyle x\equiv \pm n^{\frac {p+1}{4}}} is much faster) the binary expression of ( p + 1 ) / 2 {\displaystyle (p+1)/2} has m − 1 {\displaystyle m-1} digits, of which <i>k</i> are ones. So for computing a ( p + 1 ) / 2 {\displaystyle (p+1)/2} power of ( a + ω ) {\displaystyle \left(a+\omega \right)} , the first formula has to be used n − k − 1 {\displaystyle n-k-1} times and the second k − 1 {\displaystyle k-1} times. </p><p>For this, Cipolla's algorithm is better than the <a href="/facts/Tonelli%25E2%2580%2593Shanks_algorithm/tdxF6hfd">Tonelli–Shanks algorithm</a> if and only if S ( S − 1 ) > 8 m + 20 {\displaystyle S(S-1)>8m+20} , with 2 S {\displaystyle 2^{S}} being the maximum power of 2 which divides p − 1 {\displaystyle p-1} .<a class="footnote-ref" id="fnref:4" href="#fn:4"><sup>4</sup></a> </p> <h2 id="prime-power-moduli">Prime power moduli</h2> <p>According to Dickson's "History Of Numbers", the following formula of Cipolla will find square roots modulo powers of prime: <a class="footnote-ref" id="fnref:5" href="#fn:5"><sup>5</sup></a> <a class="footnote-ref" id="fnref:6" href="#fn:6"><sup>6</sup></a> </p> 2 − 1 q t ( ( k + k 2 − q ) s + ( k − k 2 − q ) s ) mod p λ {\displaystyle 2^{-1}q^{t}((k+{\sqrt {k^{2}-q}})^{s}+(k-{\sqrt {k^{2}-q}})^{s}){\bmod {p^{\lambda }}}} where t = ( p λ − 2 p λ − 1 + 1 ) / 2 {\displaystyle t=(p^{\lambda }-2p^{\lambda -1}+1)/2} and s = p λ − 1 ( p + 1 ) / 2 {\displaystyle s=p^{\lambda -1}(p+1)/2} where q = 10 {\displaystyle q=10} , k = 2 {\displaystyle k=2} as in this article's example <p>Taking the example in the wiki article we can see that this formula above does indeed take square roots modulo prime powers. </p><p>As </p> 10 mod 13 3 ≡ 1046 {\displaystyle {\sqrt {10}}{\bmod {13^{3}}}\equiv 1046} <p>Now solve for 2 − 1 q t {\displaystyle 2^{-1}q^{t}} via: </p> 2 − 1 10 ( 13 3 − 2 ⋅ 13 2 + 1 ) / 2 mod 13 3 ≡ 1086 {\displaystyle 2^{-1}10^{(13^{3}-2\cdot 13^{2}+1)/2}{\bmod {13^{3}}}\equiv 1086} <p>Now create the ( 2 + 2 2 − 10 ) 13 2 ⋅ 7 mod 13 3 {\displaystyle (2+{\sqrt {2^{2}-10}})^{13^{2}\cdot 7}{\bmod {13^{3}}}} and ( 2 − 2 2 − 10 ) 13 2 ⋅ 7 mod 13 3 {\displaystyle (2-{\sqrt {2^{2}-10}})^{13^{2}\cdot 7}{\bmod {13^{3}}}} (See here for mathematica code showing this above computation, remembering that something close to complex modular arithmetic is going on here) </p><p>As such: </p> ( 2 + 2 2 − 10 ) 13 2 ⋅ 7 mod 13 3 ≡ 1540 {\displaystyle (2+{\sqrt {2^{2}-10}})^{13^{2}\cdot 7}{\bmod {13^{3}}}\equiv 1540} and ( 2 − 2 2 − 10 ) 13 2 ⋅ 7 mod 13 3 ≡ 1540 {\displaystyle (2-{\sqrt {2^{2}-10}})^{13^{2}\cdot 7}{\bmod {13^{3}}}\equiv 1540} <p>and the final equation is: </p> 1086 ( 1540 + 1540 ) mod 13 3 ≡ 1046 {\displaystyle 1086(1540+1540){\bmod {13^{3}}}\equiv 1046} which is the answer. <h2 id="sources">Sources</h2> <ul><li>E. Bach, J.O. Shallit <i>Algorithmic Number Theory: Efficient algorithms</i> MIT Press, (1996)</li></ul> <h2 id="references">References</h2> <ol> <li id="fn:1"><p>Dickson, Leonard Eugene (1919). History of the Theory of Numbers. Vol. 1. Washington, Carnegie Institution of Washington. p. 218. <a href="https://archive.org/stream/historyoftheoryo01dick#page/218/mode/2up" target="_blank">https://archive.org/stream/historyoftheoryo01dick#page/218/mode/2up</a> <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li> <li id="fn:2"><p>R. Crandall, C. Pomerance Prime Numbers: A Computational Perspective Springer-Verlag, (2001) p. 157 <a href="#fnref:2" class="footnote-back-ref">↩</a></p></li> <li id="fn:3"><p>"M. Baker Cipolla's Algorithm for finding square roots mod p" (PDF). Archived from the original (PDF) on 2017-03-25. Retrieved 2011-08-24. <a href="https://web.archive.org/web/20170325174340/http://people.math.gatech.edu/~mbaker/pdf/cipolla2011.pdf" target="_blank">https://web.archive.org/web/20170325174340/http://people.math.gatech.edu/~mbaker/pdf/cipolla2011.pdf</a> <a href="#fnref:3" class="footnote-back-ref">↩</a></p></li> <li id="fn:4"><p>Tornaría, Gonzalo (2002). "Square Roots Modulo P". LATIN 2002: Theoretical Informatics. Lecture Notes in Computer Science. Vol. 2286. pp. 430–434. doi:10.1007/3-540-45995-2_38. ISBN 978-3-540-43400-9. <a href="978-3-540-43400-9" target="_blank">978-3-540-43400-9</a> <a href="#fnref:4" class="footnote-back-ref">↩</a></p></li> <li id="fn:5"><p>"History of the Theory of Numbers" Volume 1 by Leonard Eugene Dickson, p218, Chelsea Publishing 1952 read online <a href="https://archive.org/stream/historyoftheoryo01dick#page/218/mode/2up" target="_blank">https://archive.org/stream/historyoftheoryo01dick#page/218/mode/2up</a> <a href="#fnref:5" class="footnote-back-ref">↩</a></p></li> <li id="fn:6"><p>Michelle Cipolla, Rendiconto dell' Accademia delle Scienze Fisiche e Matematiche. Napoli, (3),10,1904, 144-150 <a href="#fnref:6" class="footnote-back-ref">↩</a></p></li> </ol>