If $f\left(x,y\right)$ is a quadratic form with integer coefficients, f is said to represent p if there are integers x and y such that $f\left(x,y\right)=p$. All that remains is to persuade yourself that you have verified the following theorem for a prime number $p<1000$ and different from 2 and 5.

−5 is a square modulo p if and only if p is represented by $x^2+5y^2$ or by $2x^2+2xy+3y^2$. The proof of this fact for any prime number, already stated by Fermat, goes back to Euler. There are two parts to the proof: the first part uses Legendre’s quadratic law of reciprocity: if p and q are two distinct odd primes, $\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)={\left(-1\right)}^{\left(p-1\right)\left(q-1\right)/4}$.

$\left(\frac{-5}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{5}{p}\right)={\left(-1\right)}^{\left(p-1\right)/2}\left(\frac{5}{p}\right)$

In the following, a form will be a quadratic form $a{x}^2+bxy+c{y}^2$ with a, b, and c integers prime to each other, $a>0$.

We say that a form f properly represents an integer m if there are integers u and v prime to each other such that $f\left(u,v\right)=m$. Such a solution $\left(u,v\right)$ is then said to be primitive.

Two forms $f\left(x,y\right)$ and $g\left(x,y\right)$ are said to be equivalent (or properly equivalent) if there are integers $u,v,w$ and t such that $f\left(x,y\right)=g\left(ux+vy,wx+ty\right)$ and $ut-vw=\pm 1$ ( resp. $ut-vw=1$). These are equivalence relationships.

A form $f\left(x,y\right)$ properly represents an integer m if and only if $f\left(x,y\right)$ is properly equivalent to a form of the type $m{x}^2+bxy+c{y}^2$ for b and c suitable. Indeed, suppose that there are you and v prime to each other such that $f\left(x,y\right)=m$. By Bézout’s theorem, there exists w and t such that $uw-vt=1$ and $f\left(ux+ty,vx+wy\right)$ is of the form $m{x}^2+axy+c{y}^2$. The reverse is clear.

We call the discriminant of the form $a{x}^2+bxy+c{y}^2$ the integer $D=b^2-4ac$ [7] .

Let D be an integer ≡ 0 or 1 mod 4 and m an odd integer prime to D. Then there is a form of discriminant D properly representing m if and only if D is a square modulo m.

Indeed, if there is a form properly representing m, we can take it of the type $f\left(x,y\right)=m{x}^2+bxy+c{y}^2$. Its discriminant $=b^2-4mc$ is indeed a square modulo m.

Conversely, suppose that D is a square modulo m: $D\equiv b^2\mathrm{mod}m$ with $b\in \mathbb{Z}$ Its discriminant $D=b^2-4mc$ is indeed a square modulo m: $D\equiv b^2\mathrm{mod}m$ with $b\in \mathbb{Z}$. Since m is odd, we can take b and D of the same parity (even if it means changing b to b + m). Like $D\equiv 0$ or 1 mod 4, we then have $D\equiv b^2\mathrm{mod}4m$.

Let c be the integer such that $D=b^2-4mc$. Then, $m{x}^2+bxy+c{y}^2$ properly represents m and the integers $a=m,b$ and c are prime to each other because m is prime with D. We deduce the result of the following corollary: [8] .

Let n be an integer and p an odd prime that does not divide n. Then p is represented by a form of discriminant-4m if and only if $\left(\frac{-n}{p}\right)=1$.

Let’s go back to the forms $x^2+5y^2$ and $2x^2+2xy+3y^2$ of discriminant-20. To show the following theorem, all that remains is to show that the forms of discriminant-20 are all properly equivalent to one of the two forms $x^2+5y^2$ and $2x^2+2xy+3y^2$.

−5 is a square modulo p if and only if p is represented by par $x^2+5y^2$ or by $2x^2+2xy+3y^2$.

An odd prime p prime to 5 is represented by $x^2+5y^2$ si and only if $p\equiv 1$ or 9 mod 20, it is represented by $2x^2+2xy+3y^2$ if and only if $p\equiv 3$ or 7 mod 20.

Let’s go back to the shape equivalence classes.

A form $a{x}^2+bxy+c{y}^2$ of a negative discriminant is said to be reduced if $\left|b\right|\le a\le c$ and $b\ge 0$ in cases where $\left|b\right|=a$ or $a=c$. There is only a finite number of reduced forms of the discriminant $D<0$ given (we have indeed $-D=4ac-b^2\ge 4a^2-a^2=3a^2$ and therefore $a\le \sqrt{\frac{-D}{3}}$, there is a finite number of a and b and therefore of c since $b^2-4ac=D$).

Any form of negative discriminant is properly equivalent to a single reduced form. Thus, the set of equivalence classes C(D) (for proper equivalence) of the given discriminant forms is finite [9] .

The cardinal h(D) of C(D) is called the number of classes of discriminant D.

The Cornacchia algorithm is good in the following cases:

1) $n=x^2+d{y}^2$ with d and n positive integers;

2) $n=a{x}^2+bxy+c{y}^2$ and $a>0$, $c>0$, $D=b^2-4ac<0$, $\left|D\right|>16$ et $\left|b\right|\le \frac{\left|D\right|-16}{8}$ and n integer ≥ 2sup(a, c).

So, Cornacchia’s algorithm is good for f and n and if you can’t find any qualifying pairs, The equation $n=f\left(x,y\right)$ has no integer solutions.

Let a be an integer such as −1 or a square mod a. If b is an integer verifying $0\le b\le a/2$ et $b^2\equiv -1\mathrm{mod}a$, the first two remainder $<\sqrt{a}$ in the algorithm of Euclidean division of a by b give a primitive solution of the equation $x^2+y^2=a$. Moreover, all primitive solutions are obtained in this way. A trivial transformation is one of the transformations $\left(x,y\right)\mapsto \left(\pm x;\pm y\right)$ ou $\left(x,y\right)\mapsto \left(\pm y;\pm x\right)$. A primitive solution of the equation is a solution $\left(x,y\right)$ with x and y primes between them [11] .

Consider the sequence of equations

$\{\begin{array}{l}{r}_0=q_1{r}_1+r_2\\ r_{i-1}=q_i{r}_i+r_{i+1}\\ r_{n-1}=q_n{r}_n+r_{n+1}\end{array}$ (4)

It is easy to see that by substitution, we can write

$r_0=f_n\left(q_1,\cdots ,q_n\right)r_n+g_{n-1}\left(q_1,\cdots ,q_{n-1}\right)r_{n+1}$ (5)

where the $f_m$ and $g_m$ are polynomials in $q_1,\cdots ,q_n$ of partial degree 1 in relation to each of the $q_i$.

Relationships

$r_0=f_{n-1}\left(q_1,\cdots ,q_{n-1}\right)r_{n-1}+g_{n-2}\left(q_1,\cdots ,q_{n-1}\right)r_n$

$r_{n-1}=q_n{r}_n+r_{n+1}$

Implies that:

$r_0=\left(f_{n-1}\left(q_1,\cdots ,q_{n-1}\right)q_n+g_{n-2}\left(q_1,\cdots ,q_{n-1}\right)\right)r_n+f_{n-1}\left(q_1,\cdots ,q_{n-1}\right)r_{n+1}$ (6)

We deduce that $g_{n-1}=f_{n-1}$ and therefore that

$r_0=f_n\left(q_1,\cdots ,q_n\right)r_n+f_{n-1}\left(q_1,\cdots ,q_{n-1}\right)r_{n+1}$ (7)

The polynomial $f_n$ has the property

$f_n\left(q_1,\cdots ,q_n\right)=f_n\left(q_n,\cdots ,q_1\right)$

Indeed, we have

$\left(\begin{array}{c}{r}_{i-1}\\ r_i\end{array}\right)=\left(\begin{array}{cc}{q}_i& 1\\ 1& 0\end{array}\right)\left(\begin{array}{c}{r}_i\\ r_{i+1}\end{array}\right)$

Thus, $f_n\left(q_1,\cdots ,q_n\right)$ is the coefficient at the top left of the matrix

$\left(\begin{array}{cc}{q}_i& 1\\ 1& 0\end{array}\right)\cdots \left(\begin{array}{cc}{q}_n& 1\\ 1& 0\end{array}\right)$

This coefficient is also that of its transpose (in the same place) which is equal to $\left(\begin{array}{cc}{q}_n& 1\\ 1& 0\end{array}\right)\cdots \left(\begin{array}{cc}{q}_i& 1\\ 1& 0\end{array}\right)$ and is $f_n\left(q_1,\cdots ,q_n\right)$. We can explicitly compute $f_1,\cdots ,f_7$ with MAPLE using the recurrence relation: $f_n=f_{n-1}{q}_n+f_{n-2}$. We find

$\{\begin{array}{l}{f}_1=q_1\\ f_2=q_2{q}_1+1\\ f_3=q_3{q}_2{q}_1+q_3+q_1\\ f_4=q_4{q}_3{q}_2{q}_1+q_4{q}_3+q_4{q}_1+q_2{q}_1+1\\ f_5=q_5{q}_4{q}_3{q}_2{q}_1+q_5{q}_4{q}_3+q_5{q}_4{q}_1+q_5{q}_2{q}_1+q_3{q}_2{q}_1+q_5+q_3+q_1\\ f_6=q_6{q}_5{q}_4{q}_3{q}_2{q}_1+q_6{q}_5{q}_4{q}_3+q_6{q}_5{q}_4{q}_1+q_6{q}_5{q}_2{q}_1+q_6{q}_3{q}_2{q}_1\\ +q_4{q}_3{q}_2{q}_1+q_6{q}_5+q_6{q}_3+q_6{q}_1+q_4{q}_3+q_4{q}_1+q_2{q}_1+1\\ f_7=q_7{q}_6{q}_5{q}_4{q}_3{q}_2{q}_1+q_7{q}_6{q}_5{q}_4{q}_3+q_7{q}_6{q}_5{q}_4{q}_1+q_7{q}_6{q}_5{q}_2{q}_1\\ +q_7{q}_6{q}_3{q}_2{q}_1+q_7{q}_4{q}_3{q}_2{q}_1+q_5{q}_4{q}_3{q}_2{q}_1+q_7{q}_6{q}_5+q_7{q}_6{q}_3\\ +q_7{q}_6{q}_1+q_7{q}_4{q}_3+q_7{q}_4{q}_1+q_7{q}_2{q}_1+q_5{q}_4{q}_3+q_5{q}_4{q}_1\\ +q_5{q}_2{q}_1+q_3{q}_2{q}_1+q_7+q_5+q_5+q_1\end{array}$ [12]

It will be noted that the monomials involved in $f_n$ are exactly those obtained by removing from $q_1,\cdots ,q_n$ a certain number of successive pairs of elements $\left(q_i,\cdots ,q_{i+1}\right)$.

If $1<m<n$, we have

$r_m=f_{n-m}\left(q_{m+1},\cdots ,q_n\right)r_n+f_{n-m-1}\left(q_{m+1},\cdots ,q_{n-1}\right)r_{n+1}$

$r_{m+1}=f_{n-m-1}\left(q_{m+2},\cdots ,q_n\right)r_n+f_{n-m-2}\left(q_{m+2},\cdots ,q_{n-1}\right)r_{n+1}$ (8)

$r_0=f_m\left(q_1,\cdots ,q_m\right)r_m+f_{m-1}\left(q_1,\cdots ,q_{m-1}\right)r_{m+1}$

It is easy to deduce that:

$\begin{array}{l}{f}_n\left(q_1,\cdots ,q_n\right)\\ =f_m\left(q_1,\cdots ,q_m\right)f_{n-m}\left(q_{m+1},\cdots ,q_n\right)+f_{m-1}\left(q_1,\cdots ,q_{m-1}\right)f_{n-m-1}\left(q_{m+2},\cdots ,q_n\right)\end{array}$ (9)

Particular case

Suppose that n is even and that $\left(q_1,\cdots ,q_n\right)=\left(q_n,\cdots ,q_1\right)$, that is, that is, $q_i=q_{n+1-i}$.

For $m=\frac{n}{2}$, the equation becomes:

$\begin{array}{l}{f}_n\left(q_1,\cdots ,q_n\right)\\ =f_{\frac{n}{2}}\left(q_1,\cdots ,q_{\frac{n}{2}}\right)f_{\frac{n}{2}}\left(q_{\frac{n}{2}+1},\cdots ,q_n\right)+f_{\frac{n}{2}-1}\left(q_1,\cdots ,q_{\frac{n}{2}-1}\right)f_{\frac{n}{2}-1}\left(q_{\frac{n}{2}+2},\cdots ,q_n\right)\end{array}$

$f_n\left(q_1,\cdots ,q_n\right)=f_{\frac{n}{2}}{\left(q_1,\cdots ,q_{\frac{n}{2}}\right)}^2+f_{\frac{n}{2}-1}{\left(q_1,\cdots ,q_{\frac{n}{2}-1}\right)}^2=f_{\frac{n}{2}}{\left(q_n,\cdots ,q_{\frac{n}{2}+2}\right)}^2$

Either $f_n\left(q_1,\cdots ,q_n\right)=f_{\frac{n}{2}}{\left(q_{\frac{n}{2}+1},\cdots ,q_n\right)}^2+f_{\frac{n}{2}-1}{\left(q_{\frac{n}{2}+2},\cdots ,q_n\right)}^2$

Let a and b now be two positive integers prime to each other with $b<a$. The sequence of the quotients associated with $\left(a,b\right)$ is the sequence of successive quotients obtained by applying Euclid’s algorithm to a and b: we set $r_0=a$, $r_1=b$ and then we define by recurrence

$q_i=\left[r_{i-1}/r_i\right]\text{et}{r}_{i+1}=r_{i-1}{q}_i{r}_i$(10)

We denote n the integer such that $r_n=pgcd\left(a,b\right)=1$ and we call it the length of the algorithm. So we have $r_{i-1}=0$.

We define another sequence of integers $t_i$ by:

$t_0=0,t_1=1\text{et}{t}_{i+1}=t_{i-1}-q_i{t}_i\text{by}i=1,\cdots ,n+1$

We then have the ties:

$r_i=s_{i}a+t_{i}b$

If $s_i$ is the sequence defined by the same recurrence relation and first terms 1 and 0. By recurrence, it is easy to see that the $t_i$ are of alternating sign, more precisely $\left|t_i\right|={\left(-1\right)}^{i-1}{t}_i$.

Hence the relationship

$\left|t_{i+1}\right|=\left|t_{i-1}\right|+q_i\left|t_i\right|$(11)

The rest of the $\left|t_i\right|$ is therefore increasing. Thus, we can see the $\left|t_{i+1-i}\right|$ as the sequence of successive remainders in Euclid’s algorithm applied to $\left|t_{i+1}\right|$ and $\left|t_i\right|$. Since $r_{i+1}=0$ and a and b are prime to each other, for $i=n+1$ indicates that a divides $r_{n+1}=0$. We actually have $\left|t_{n+1}\right|=a$ because $\left|t_{n+i}\right|=f_n\left(q_n,\cdots ,q_1\right)=f_n\left(q_1,\cdots ,q_n\right)=a$ What can we say now about $\left|t_n\right|$? we have $t_n<a$ et $b{t}_n\equiv 1\mathrm{mod}a$.

Let a and b be two integers with $b<a/2$. The sequence of successive quotients in Euclid’s algorithm (of length n) starting with a and b is symmetric if and only if $b^2\equiv \pm 1\mathrm{mod}a$ and we then have $b^2\equiv {\left(-1\right)}^{n+1}\mathrm{mod}a$ [13] .

Let’s prove the lemma. We have just seen that the sequence of $\left|t_i\right|$ is increasing, that the $t_i$ are of alternating sign, with $t_n$ of the sign of ${\left(-1\right)}^{n+1}$, and that $\left|t_{i+1}\right|=a$.

Suppose $b^2\equiv \epsilon \mathrm{mod}a$ with $\epsilon =\pm 1$.

Show that $\left|t_n\right|=\epsilon b$. We notice that $t_{n}b\equiv 1\mathrm{mod}a$. Like $b^2\equiv \epsilon \mathrm{mod}a$, we deduce that $t_n\equiv \epsilon b\mathrm{mod}a$. Moreover, $\left|t_n\right|<\left|t_{n+1}\right|=a$, therefore $\left|t_n\right|=\epsilon b$ where $\left|t_n\right|=\epsilon \left(b-a\right)$.

Suppose $t_n=\epsilon \left(b-a\right)$. Wich means $\left|t_n\right|=a-b$. The euclidian division of $\left|t_{n+i}\right|=a$ par $\left|t_n\right|=a-b$ gives

$\left|t_{n+i}\right|=\left|t_n\right|+b$

Since $b<a-b$. So $\left|t_{n+i}\right|=b$ and we have

$a=\left(a-b\right)+\left|t_{n-i}\right|$

$a-b=q_{n-1}b+\left|t_{n-2}\right|$

$b=q_{n-2}\left|t_{n-2}\right|+\left|t_{n-3}\right|$

From where

$a=\left(q_{n-1}+1\right)b+\left|t_{n-2}\right|$

$b=q_{n-2}\left|t_{n-2}\right|+\left|t_{n-3}\right|$

Euclid’s algorithm of a by b would be of length $n-1$. Contradiction, so $t_n=\epsilon b$. It is then clear that the sequences $\left(q_1,\cdots ,q_n\right)$ and $\left(q_n,\cdots ,q_1\right)$ are equa. In addition, we have $\epsilon ={\left(-1\right)}^{n+1}$. Let us now suppose the sequences $\left(q_1,\cdots ,q_n\right)$ and $\left(q_n,\cdots ,q_1\right)$ equal. So we have $\left|t_{n+1}\right|=a$ and $t_n={\left(-1\right)}^{n+1}b$. We also always have $t_{n}b\equiv 1\mathrm{mod}a$ which implies that $b^2\equiv {\left(-1\right)}^{n+1}\mathrm{mod}a$ and ends the proof of the lemma. Let’s use the above to prove the theorem. A necessary condition for a primitive solution of the equation $x^2+y^2=a$ to exist is that 4 does not divide a and that all odd primes dividing a are congruent to 1 mod 4 (this is a condition for −1 to be a square mod a). We now assume so. There is then an integer $b<\frac{a}{2}$ such that $b^2\equiv -1\mathrm{mod}a$; Let’s choose one.

By applying

$f_n\left(q_1,\cdots ,q_n\right)=f_{\frac{n}{2}}{\left(q_{\frac{n}{2}+1},\cdots ,q_n\right)}^2+f_{\frac{n}{2}-1}{\left(q_{\frac{n}{2}+2},\cdots ,q_n\right)}^2$ and by denoting $r_i$ the successive remainders as before, we obtain that the roots $r_{\frac{n}{2}}$ and $r_{\frac{n}{2}+1}$ check

$a=r_{\frac{n}{2}}^2+r_{\frac{n}{2}+1}^2$

Although of course $0\le r_{\frac{n}{2}}\le \sqrt{a}$.

Let us show that $r_{\frac{n}{2}}$ is the first remainder less than $\sqrt{a}$. Let us first note that for any integer i, we have

$r_i^2+r_{n+1-i}^2\equiv 0\mathrm{mod}a$

Indeed, as $t_{i}b\equiv r_i\mathrm{mod}a$ and $b^2\equiv -1\mathrm{mod}a$, by squaring, $t_i^2+r_i^2\equiv \mathrm{mod}a$; since $t_i=r_{n+1-i}$, the congruence can be deduced from this. Suppose that there exists $m<\frac{n}{2}$ such that $r_m\le \sqrt{a}$. Then, $r_{n+1-m}^2\le a$ and we necessarily have $r_m^2+r_{n+1-m}^2=a$. On the other hand, $r_m/r_{n+1-m}^2\equiv b\mathrm{mod}a\equiv r_{\frac{n}{2}}/r_{\frac{n}{2}+1}$. All that remains is to show that with the nearest transformation by $\left(x,y\right)\mapsto \left(\pm x,\pm y\right)$ and $\left(x,y\right)\mapsto \left(\pm y,\pm x\right)$, there is only one solution $\left(x,y\right)$ to the equation $x^2+y^2=a$ of quotient x/y mod has given, which will imply that $r_m=r_{\frac{n}{2}-1}$ and therefore $m=\frac{n}{2}-1$. To do this, let’s take two pairs $\left(x,y\right)$ and $\left(x^{\prime },y^{\prime }\right)$ with $x<y$, $x^{\prime }<y^{\prime }$ et $x/y\equiv x^{\prime }/y^{\prime }\mathrm{mod}a$. So we have the equations $x^2+y^2=a$, ${x^{\prime }}^2+{y^{\prime }}^2=a$ and $x{y}^{\prime }-x^{\prime }y=ar$, for r an integer. Let us eliminate $x^{\prime }$ in the second equation. We easily obtain the equation ${y^{\prime }}^2-2rx{y}^{\prime }+a{r}^2-y^2=0$ whose discriminant must be positive (and even a square in $\mathbb{Z}$); it is $\left(1-r^2\right){\beta }^2$. We therefore necessarily have $r=0$, or $r=\pm 1$. For $r=0$, we get $y^2={y^{\prime }}^2$, for $r=\pm 1$, we get $y^{\prime }\pm x=0$, which proves our assertion. In conclusion, we have shown that the different choices of b verifying $0<b<a/2$ and $b^2\equiv -1\mathrm{mod}a$ give the different primitive solutions of the equation $x^2+y^2=a$, with the nearest transformation by $\left(x,y\right)\mapsto \left(\pm x,\pm y\right)$ and $\left(x,y\right)\mapsto \left(\pm y,\pm x\right)$, in the following way: the solution is given by the first pair of remainders less than $\sqrt{a}$ in Euclid’s algorithm of a by such a b. For an odd divisible only by prime numbers congruent to à 1 mod 4, we find $4\cdot 2^{\lambda }$ solutions, where $\lambda$ is the number of prime factors in the decomposition of n (if p is odd, −1 is a square in $\mathbb{Z}/p^S\mathbb{Z}$ if and only if $p\equiv 1\mathrm{mod}4$ and then there are exactly two solutions; then use the Chinese remainder theorem).

In any case, Cornacchia’s algorithm provides a systematic way to find integer solutions to quadratic equations, exploiting the particular properties of each quadratic form. It is a very powerful tool in number theory. In any case, Cornacchia’s algorithm provides a systematic way to find integer solutions to quadratic equations, exploiting the particular properties of each quadratic form. It is a very powerful tool in number theory.

The Cornacchia algorithm is an efficient method for solving the Pell-Fermat equation, which has the following form: $x^2-d{y}^2=1$ where D is a positive non-square integer. The computational complexity of the adapted Cornacchia algorithm depends mainly on two factors:

1) The size of the numbers involved in the calculation:

2) The number of iterations needed to find a solution:

In conclusion, the complexity of the adapted Cornacchia algorithm can be characterized as follows:

The actual complexity therefore depends heavily on the value of D and the size of the numbers involved. For relatively small D values, the adapted Cornacchia algorithm is usually very efficient and can be used to solve reasonably sized Pell-Fermat equations [14] .

Let’s go through a comparison between the adapted Cornacchia algorithm and the original one.

In summary, the adapted Cornacchia algorithm is generally more efficient and more convenient to implement than the original algorithm, especially for very large D values. However, for relatively small D values, the two algorithms have similar performance.

1) Verify that the values of x and y are positive integers.

2) Calculate the product $x^2-D{y}^2$ and check that it is equal to 1.

3) Compare the candidate solution to already known solutions to identify the smallest in value.

4) Make sure that there are no other primitive solutions for the same value of D. By following these criteria, it can be guaranteed that the solution identified is the unique and fundamental primitive solution of the Pell-Fermat equation for the value of D considered.

The range of possible solutions for the equations of the quadratic form $x^2–D{y}^2=1$ (Pell-Fermat equation) depends on several factors.

1) Valeur of D

2) Sign of solutions

3) Solution size

1) Ineligible D values.

2) Numerical precision problems.

To overcome these problems, it may be necessary to use arbitrarily precise calculation libraries or to adapt the algorithm to better handle very large values.

In summary: the range of possible solutions for the Pell-Fermat equations is very wide and depends mainly on the value of D. however, the adapted Cornacchia algorithm may encounter limitations when values become very large, requiring adaptations to ensure its robustness.

1) Quadratic generating forms:

2) Cubic generating forms.

3) Higher-order generative forms.

1) Number theory and number geometry.

2) Diophantine équations

3) Additive number theory.

4) Cryptography and number theory.