DIOPHANTINE EQUATIONS x⁴ + y⁴ = zⁿ, n ≥ 0 APPLICATION IN CRYPTOGRAPHY René-Louis Clerc (june 2026) (*)

- ABSTRACT
We provide an overview of the known and expressible solutions in the family of Diophantine quartic equations
  x⁴ + y⁴ = zⁿ, xyz ≠ 0, n ≥ 0.
We will determine the values of n that lead to non-trivial solutions and describe families of solutions that are not necessarily primitive and belong essentially to Z.
For the case x = y with odd exponents n greater than 1, we will express all possible solutions parametrically.
Finally, we will describe an application to asymmetric cryptography by considering the finite field Z/pZ, where p is a large prime number satisfying two conditions, and by defining a one-way function associated with these equations.

-EQUATIONS DIOPHANTIENNES x⁴ + y⁴ = zⁿ, n ≥ 0
APPLICATION EN CRYPTOGRAPHIE
Nous proposons un aperçu des solutions connues et exprimables de la famille des équations quartiques diophantiennes
  x⁴ + y⁴ = zⁿ, xyz ≠ 0, n ≥ 0.
Nous déterminerons les valeurs de n qui conduisent à des solutions non triviales et expliciterons des familles de solutions non nécessairement primitives appartenant essentiellement à Z.
Pour le cas x = y avec des exposants n impairs supérieurs à 1, nous exprimerons paramétriquement toutes les solutions possibles.
Nous décrirons enfin un schéma d'application à la cryptographie asymétrique en nous plaçant dans le corps fini Z/pZ, p étant un grand nombre premier vérifiant deux conditions, en définissant une fonction à sens unique associée à ces dernières équations.

- Mathematics Subject Classification-MSC2020: 11D41, 11D45, 11J25.
- Keywords: Diophantine equations, Fermat's equation, solutions of Diophantine equations, modular arithmetic.

- INTRODUCTION
Many Diophantine equations have been studied ([2], [3], [4], [5], [6], [7], [8], [10], [11], [12], [13], [14], [15], [16], [17], [18], ...) connected with the one in the title.
For example, K. Gyory studied in ([1],[2]) the Diophantine equation x^p + y^p = cz^p and in [5] B. J. Powell proved that this equation has no integer solutions for special values of p.
Here we will consider the equations given in the title for various values of the positive integer exponent n.
For our quartic equations (of the Fermat–Catalan type)
  x⁴ + y⁴ = zⁿ,
we recall that their primitive solutions are defined by the conditions
  xyz ≠ 0 and gcd(x,y,z) = 1.
However, we will consider examining and listing all possible non-trivial solutions for a given n, especially since, as soon as n > 2, we know that Beal’s conjecture ([11]), which remains neither proven nor disproven, states that these equations have no non-trivial primitive integer solutions.
It should be noted that the parity of the exponent 4 of the variables x and y allows us to immediately extend the set of solutions from N to Z: if the pair (x, y) is a solution, then the combinations (±x, ±y) are also solutions.

-1- Case n = 0; x⁴ + y⁴ = 1
There is no non-trivial solution.
In R², the set of solutions forms a closed curve that is symmetric about the axes and the origin and resembles a square with rounded corners. If we replace the exponent 4 with an even number m > 1, our curve is one of the intermediate superellipses between the circle for m = 2 and the square for m → ∞.

- 2- Case n = 1; x⁴ + y⁴ = z
This is the classic Waring's problem (1770): there are infinitely many integers z that can be expressed as a sum of two powers of four (see 2quatre, A003336):
2, 17, 32, 82, 97, 162, 257, 272, 337, 512, 626, 641, 706, 881, 1250, 1297, 1312, 1377, 1552, 1921, 2402, 2417, 2482, 2592, 2657, 3026, 3697, 4097, 4112, 4177, 4352, 4721, 4802, 5392, 6497, 6562, 6577, 6642, 6817, 7186, 7857, 8192, 8962, 10001, 10016, 10081, 10256, 10625 ...
For example, 1 312 = 2⁴ + 6⁴, 10 081 = 3⁴ + 10⁴, 71 769 617 = 19⁴ + 92⁴.
The smallest number with two solutions is 635 318 657 = 59⁴ + 158⁴ = 133⁴ + 134⁴ (A018786), the next is 3 262 811 042 = 7⁴ + 239⁴ = 157⁴ + 227⁴ (A003824).
(59, 158) is a so-called primitive solution (gcd(59, 158, 635 318 657) = 1), whereas (118, 316) is not primitive because: 118⁴ + 316⁴ = 16*(59⁴ + 158⁴).
Note that there are no cases with 3 solutions (see Bill Butler: there are 1 413 integers corresponding to two primitive solutions).
https://www.durangobill.com/RamanujanPics/Rama4thPower.html
- 3- Case n = 2; x⁴ + y⁴ = z²
There is no non-trivial integer solution according to Fermat and Kummer.
On the other hand, there are infinitely many non-integer solutions, such as (2, 2, 4√2), (2, 3, √97), (1, 4, √257) or (X, Y, √(X⁴ + Y⁴)) for all X, Y.

- 4- Case n = 4; x⁴ + y⁴ = z⁴
According to Fermat–Wiles’ theorem [9], there are no non-trivial integer solutions.
Here, too, there are infinitely many non-integer solutions.

- 5- Case n = 2p; x⁴ + y⁴ = z^2p , p positif
These equations reduce to Fermat equations
  (x²)² + (y²)² = (z^p)²
It can be shown quite easily that there is no non-trivial integer solution for all p ≥ 1.
For any p, there will always be an infinite number of non-integer solutions.
Following Fermat and Euler, assume a minimal solution, transform the equation with another smaller solution to arrive at a contradiction (method of infinite descent).

- 6- Case n = 2p + 1; x⁴ + y⁴ = z^2p+1 , p > 0
Interesting cases ([1], [8]) corresponding to odd n greater than 1: there are non-trivial integer solutions for every p, at least in the case where x = y.
-Property 1

For any p > 0, all integer solutions of 2x4 = z2p+1 will necessarily be of the general form x = An2k(n), n = 2p + 1, where the expression for k depends on n and A is an integer, which may be relative; z will then be of the form z = A42m(n).

Let's first look for solutions where x = y = 2^k.
This gives 1 + 4k ≡ 0 (mod n), so we obtain k(n), which will allow us to express the general form of all integer solutions x = y for the case n = 2p+1 under consideration.
Since k(n) is a solution to 1 + 4k = mn for m an integer, the minimum values of k and m can be easily calculated in terms of n (see the following paragraphs):
  n = 3, k = 2, m = 3;
  n = 5, k = 1, m = 1;
 ....
  n = 15, k = 11, m = 3.
This will yield all non-trivial solutions where x = y. Indeed, given a solution (x, z), for any prime number p that divides x:
1) - If p > 2, the exponent of p in 2x⁴ is 4v_p(x) and in z it is nv_p(z). The equality 4v_p(x) = nv_p(z) and gcd(4,n) = 1 implies that v_p(x) must be a multiple of n, hence the factor Aⁿ.
2) - If p = 2, the exponent of 2 in 2x⁴ is 4v₂(x) + 1; it must be equal to nv₂(z), so it is exactly k(n).
In both of these cases, we can easily deduce the exponent m(n) in z (see Property 2 of paragraph 14 for the explicit form of the solutions (x, z) for all p).
We thus parameterize the infinite set of all non-trivial solutions with x = y.
We will now examine in detail some of these odd cases where n ≥ 3, by expanding the expressions for the exponents k(n) and m(n) for each n.

- 7- Case n = 3; x⁴ + y⁴ = z³
This case defines elliptic curves which may have a positive rank, and thus an infinite number of rational points and consequently an infinite number of integer solutions ([1]).
There are an infinite number of solutions x = y and none with x and y not proportional.
For any odd n > 1, integer solutions of the form x = y will necessarily be of the general form Aⁿ2^k(n) where the expression for k depends on n .
The search for solutions of the form x = y = 2^k leads to
  1 + 4k ≡ 0 (mod 3), so k = 2 + 3t.
Thus all integer solutions x = y are of the form
  x = y = 2^2+3t Π_i=1^r p_i^3u_i,
  z = 2^3+4t Π_i=1^r p_i^4u_i,
t, r, u_i integers ≥ 0, p_i any odd primes.
Examples:
  t = 0, r = 0, x = y = 4, z = 8;
  t = 0, r = 1, p₁ = 3, u₁ = 1, x = y = 108, z = 648;
  t = 1, r = 0, x = y = 32, z = 128;
  t = 0, r = 1, p₁ = 5, u₁ = 1, x = y = 500, z = 5 000.
We can find solutions where x and y are different but satisfy y = kx:
  x = (1+k⁴)²t³, y = k(1+k⁴)²t³,   z = (1+k⁴)³t⁴.
Examples:
  k = 2, t = 1, x = 289, y = 578, z = 4 913;
  k = 2, t = 2, x = 2 312, y = 4 624, z = 78 608;
  x⁴ + y⁴ = 485 735 942 131 712;
  k = 3, t = 1, x = 6 724, y = 20 172, z = 551 368;
  x⁴ + y⁴ = 167 619 550 409 708 032;
  k = 5, t = 1, x = 391 876, y = 1 959 380, z = 245 314 376;
  x⁴ + y⁴ = 14 762 808 930 988 484 877 349 376.
Are there any others?
Probably not of a different type from kx (conjecture).

- 8- Case n = 5; x⁴ + y⁴ = z⁵
This is a rather unusual and interesting case, as there are solutions x and y that are NOT proportional.
The associated curve is an elliptic curve of positive rank, hence there are infinitely many rational points and therefore infinitely many families of integer solutions.
As mentioned above, if we look for solutions of the form x = y = 2^k, we obtain k = 1 + 5t and thus all integer solutions x = y are of the form
  x = y = 2^1+5t Π_i=1^r p_i^5u_i,
  z = 2^1+4t Π_i=1^r p_i^4u_i,
t, r, u_i integers ≥ 0, p_i odd primes.
Examples:
  t = 0, r = 0, x = y = 2, z = 2;
  t = 1, r = 0, x = y = 64, z = 32;
  t = 0, r = 1, p₁ = 3, u₁ = 1, x = y = 486, z = 162;
  t = 0, r = 1, p₁ = 7, u₁ = 1, x = y = 33 614, z = 4 802;
  t = 2, r = 1, p₁ = 5, u₁ = 1, x = y = 6 400 000, z = 320 000.
It is also fairly easy to find solutions where x ≠ y but y = kx (where k is a positive integer > 1):
  x = (1+k⁴)t⁵, y = k(1+k⁴)t⁵,
  z = (1+k⁴)t⁴,
  k (>1) and t are positive integers.
Examples with k = 2:
  t = 1: 17, 34, 17;
  t = 2: 544, 1 088, 272;
  t = 3: 4 131, 8 262, 1 377;
  t = 4: 17 408, 34 816, 4 352;
  t = 5: 53 125, 106 250, 10 625.
Let us now try to construct solutions x ≠ y where y/x is rational.
If we let gcd(x,y) = d, we can write
  x = dm, y = dn, gcd(m, n) = 1, m and n positive integers;
by setting S = m⁴ + n⁴, our equation becomes d⁴S = z⁵.
By factoring S and d into prime factors, we can fairly easily express d and z in the form
  d = S^1+5vw^5u, z = S^1+4vw^4u, where u, v, and w are integers (w ≠ 0),
and derive the expressions for x and y.
Since attempts to find alternative solutions have failed, we can reasonably propose the following conjecture.
- Conjecture

All distinct integer solutions (y/x = n/m any rational number) of x4 + y4 = z5 are of the form   x = m(m4+n4)1+5vw5u, y = n(m4+n4)1+5vw5u,   z = (m4+n4)1+4vw4u, m ≠ n, n, w non-negative integers, u and v integers.

Examples:
  m = 4, n = 7, v = 0, w = 1, u = 0,
  x = 10.628, y = 18.599, z = 2.657;
  x⁴ + y⁴ = z⁵ = 132 421 277 116 505 057;
  m = 2, n = 3, v = 1, w = 2, u = 1,
  x = 53 310 208 315 456, y = 79 965 312 473 184, z = 137 397 444 112,
  x⁴ + y⁴ = z⁵ = 48 965 846 853 680 836 650 881 544 765 598 622 770 576 250 993 961 336 832.
We have shown here an infinite family of non-trivial solutions, where y/x is any rational number, but these are not primitive, since gcd(x,y,z) ≥ m⁴+n⁴ > 1.
At present, no non-trivial primitive solution is known.
It should be noted that this case is particularly interesting, even exceptional, primarily because the associated curve is elliptic of strictly positive rank.

-9- Case n = 7; x⁴ + y⁴ = z⁷
In this hyperbolic case, the absence of non-proportionnal non-trivial solution follows from the incompatibility between the Galois representation attached to the Frey curve and the space of modular forms of the corresponding level ([9]).
As in the cases where n = 3 and 5, we can express all integer solutions x = y in the form
  x = y = 2^5+7t Π_i=1^r p_i^7u_i,
  z = 2^3+4t Π_i=1^r p_i^4u_i,
t, r, u_i integers ≥ 0, p_i distinct odd primes.
Examples:
  t = 0, r = 0, x = y = 32, z = 8;
  t = 1, r = 0, x = y = 4 096, z = 128;
  t = 2, r = 0, x = y = 2¹⁹, z = 2¹¹;
  t = 0, r = 1, p₁ = 3, u₁ = 1, x = y = 69 984, z = 648;
  t = 1, r = 1, p₁ = 5, u₁ = 2, x = y = 25 000 000 000 000, z = 50 000 000.
There is no solution where x is different from y.

-10- Case n = 9; x⁴ + y⁴ = z⁹
The case n = 9 can be reduced to the cubic case by setting Z = z³ (any solution for n = 9 yields a solution for n = 3).
The latter defines an elliptic curve of rank 0, implying that only trivial non-proportionnal solutions exist.
Although no non-trivial non-proportional integer solutions are known, it is possible to obtain non-trivial proportional solutions.
There are only two “small” solutions less than à 10³:
  x = y = 4, z = 2 and x = 289, y = 578, z = 17;
corresponding solutions for n = 3: (4, 4, 8) and (289, 578, 4 913).
Note that the second solution is such that x = 2*y.
We can also obtain a solution such as y = 4x/3
  x = 340 707, y = 454 276, z = 337
Trivial rational solutions can be expressed
  x = 0, y = t⁹, z = t⁴ ou x = t⁹, y = 0, z = t⁴ for any t ∈ Q.
As mentioned above, all non-trivial integer solutions x = y are of the form
  x = y = 2^2+9t Π_i=1^r p_i^9u_i,
  z = 2^1+4t Π_i=1^r p_i^4u_i,
t, r, u_i integers ≥ 0, p_i distinct odd primes.
Here (since n = 3²) there are solutions where x is different from y.

-11- Case n = 11; x⁴ + y⁴ = z¹¹
In this hyperbolic case since 11 is prime, there is no reduction to a lower possible case (unlike in the case of n = 9).
All integer solutions where x = y are of the form
  x = y = 2^8+11t Π_i=1^r p_i^11u_i,
  z = 2^3+4t Π_i=1^r p_i^4u_i,
t, r, u_i integers ≥ 0, p_i distinct odd primes.
The small solution with m = 0 is (256, 256, 8).
There is no solution where x is different from y.

-12- Case n = 13; x⁴ + y⁴ = z¹³
With the prime 13 as the exponent, we see the same behaviour here as in the case where n = 11.
All integer solutions where x = y are of the form
  x = y = 2^3+13t Π_i=1^r p_i^13u_i,
  z = 2^1+4t Π_i=1^r p_i^4u_i,
t, r, u_i integers ≥ 0, p_i distinct odd primes.
The small solution with m = 0 is (8, 8, 2).
There is no solution where x is different from y.

-13- Case n = 15; x⁴ + y⁴ = z¹⁵
This case is not unusual, even though 15 = 3 × 5: no reduction is possible.
All integer solutions where x = y are of the form
  x = y = 2^11+15tΠ_i=1^r p_i^15u_i,
  z = 2^3+4t Π_i=1^r p_i^4u_i,
t, r, u_i integers ≥ 0, p_i distinct odd primes.
The small solution with m = 0 is (2 048, 2 048, 8).
There is no solution where x is different from y.

-14- General case: 2x⁴ = z^2p+1
In contrast to the preceding paragraphs, we will use more symmetric notation for x and z here (k will become a and m will become b).
For any positive p, you can always find solutions such as x = 2^α , z = 2^β; they must check 4α + 1 = 2βp + β.
Since, for any positive p, there are always an infinite number of solutions (α, β), by calling (a, b) the smallest ones, we can denote them as
  α = a + (2p + 1)t, β = b +4t, for any t ≥ 0.
-Property 2

For any p > 0, all integer solutions of 2x4 = z2p+1 are of the form   x = y = 2a + (2p + 1)tΠi=1r pi(2p+1)ui,   z = 2b+4t Πi=1r pi4ui, t, r, ui integers ≥ 0, pi distinct odd primes.

For any n = 2p + 1, we can easily find the pair (a, b) and thus the general form of the solutions (for example, for n = 17 we obtain a = 4 and b = 1, for n = 19, a = 14 and b = 3, ...).

-15- General case: x^p + y^p = zⁿ
Let us consider the general problem of the type
  x^p + y^p = zⁿ,
  gcd(x, y) = 1, xy ≠ 0, n and p positive integers > 2.
Let’s consider the characteristic
  q = 1/p + 1/p + 1/n = 2/p + 1/n;
the fundamental classification of pairs (p, n) will lead to
  q > 1, elliptic cases, of which there are relatively few, such as (2, 2) or (2, 3), with many possible solutions;
  q = 1, parabolic cases, such as (3, 3), which behave in a tricky way;
  q < 1, hyperbolic cases, which are by far the most common and usually have a finite number of primitive solutions ([7], [8]).
This exceptional case corresponds to the prime 5 (the first pentagonal number greater than 1) with its many and varied symbolic interpretations, such as the number of Aphrodite, the number of life, the number of material existence or the number of the five senses (see "The Symbolism of Numbers" by R. Berteaux, 2016).
We could have considered certain mixed equations of the form x^p + y^r = zⁿ, where p and r are different, in particular, the case of three exponents greater than 2 and the famous Beal conjecture ([11]) (if there are positive integer solutions, then x, y, and z have a common prime factor). As a reminder, the next Beal Prize will be awarded in 2028 ...

-16- Application in cryptography
The structure of the solutions to our quartic Diophantine equations for the symmetric case x = y with odd exponents n greater than 1 exhibits remarkable arithmetic properties that can be exploited in asymmetric cryptography.
This will lead us to the design of a one-way function.
We can use the previous results to create functions that are easy to compute in one direction but difficult to reverse. This is the technology behind zero-knowledge proofs, which are booming with the rise of cryptocurrency blockchains.
The idea is to prove that we know the solution to a complex equation (one of our equations) without ever revealing the solution itself, but by indicating that it possesses a certain arithmetic property (see Paillier's asymmetric encryption or the Diffie-Hellman symmetric protocol involving Alice and Bob).
Using the canonical projection
  π : Z → Z/pZ,
where p is a prime number (which will be chosen to be very large, on the order of 2048 bits, i.e., with more than 600 digits), we transpose our equations 2x⁴ = zⁿ from Z into the finite field Z/pZ; thus we will define a transformation function f such that
  f(x) ≡ 2x⁴ (mod p).
Any exact solution to the equation in Z automatically translates into a modular solution in Z/pZ, although the reverse is not true, of course, which provides cryptographic security.
To ensure the robustness and injectivity of the system, the following conditions must hold:
1) - n odd > 1, to obtain the solutions x = Aⁿ2^k(n) explained above;
2) - p such that gcd(n, p-1) = 1, to ensure that for every image C = f(x), there exists a unique z satisfying the relation
  zⁿ ≡ C (mod p),
that is, n is invertible modulo p-1.
The notation C is inspired by the classic RSA asymmetric encryption (algorithm defined in 1977).
Given p-1, we can compute the inverse of n modulo p-1, denoted by d ≡ 1/n (mod p-1), and thus obtain, using Fermat's Little Theorem, z from C (see proof (+))
  z ≡ C^d (mod p).
Note that z ∈ {0, 1, ..., p-1}.
The security of this function relies on the asymmetry of the computation: while evaluating f(x) is simple and fast using the binary exponentiation algorithm (for our odd n, we will write xⁿ = x*x^n-1 to speed up the calculation), its inversion requires the very difficult and tedious calculation (which is an understatement for very large p) of modular quartic roots to obtain x or modular nth roots to obtain z.
To ensure the uniqueness of the result, we prefer to find z.
Public data: n, a very large prime number p and C.
Problem: Find z such that 2x⁴ ≡ zⁿ (mod p).
We consider C ≡ 2x⁴ (mod p) to be the encrypted message (from the secret x), d to be the private key, and z to be the result (or proof value).
For a solution (x, z), the chosen value of p must be (at least) strictly greater than z in order to ensure the injectivity of the modular process.
Finding z without knowing d would correspond to the difficult problem of nth roots or discrete logarithms.
The holder of the secret x can generate a proof z (such that zⁿ ≡ 2x⁴ mod p). Using the key d, this z can be computed in polynomial time and is protected by the difficulty of nth root extraction for someone who does not know the factorization of p-1.
Note that in our approach, where starting from C = 2x⁴ we aim to ensure that C ≡ zⁿ (mod p), C can be unlocked in two ways: either via the fourth root or via the nth root.
We thus have a two-input system, where x can be considered the key of user A, z the key of user B, and C the shared secret (meeting point).
- Example 1:
n = 3 with the solution (x = 4, z = 8) and p = 107,
C = 2x⁴ = 512, so C ≡ 84 (mod 107),
gcd(n, p-1) = gcd(3, 106) = 1; we obtain d ≡ 1/3 (mod 106) ≡ 71.
z = 84^71 ≡ 8 (mod 107) .
We will give n, p, C ; find z modulo 107.
- Example 2:
n = 5 with the solution (x = 33 614, z = 4 802) and p = 20 147,
C = 2x⁴ = 2 553 352 521 523 584 032, so C ≡ 14 257 (mod 20 147) ,
gcd(n, p-1) = gcd(5, 20 146) = 1; we obtain d ≡ 1/5 (mod 20 146) ≡ 16 117.
z = 14 257^16 117 ≡ 4 802 (mod 20 147).
We will give n, p, C ; find z modulo 20 147.
With numbers of 600 digits or more, these calculations are currently impossible.
Let's take an example with larger numbers and perform the coding and calculations using Pari/Gp, a free software package specialized in number theory.
- Example 3
n = 5, x = 6 400 000, z = 320 000, p = 100 000 000 019, gcd(5, 100 000 000 018) = 1.
As mentioned above, we must perform the following calculations (Mod(C, p) is a function from Pari/Gp that uses a fast exponentiation algorithm; note that here this function will perform only about sixty multiplications instead of 60 billion, because at each step of the calculation the result is reduced modulo p, hence the extreme speed):
  C = Mod(2*x^4, p); d = Mod(1/5, p-1); Mod(C, p)^d = Mod(z, p).
We obtain, respectively:
Mod(2*6400000^4,100000000019) = Mod(92121131517, 100000000019),
Mod(1/5, 100000000018) = Mod(60000000011, 100000000018),
Mod(92121131517, 100000000019)^60000000011 = Mod(320000, 100000000019).
For the secret number x, with the public identifier C = 92 121 131 517, the prime p = 100 000 000 019, and the encryption key d = 60 000 000 011, the result obtained is indeed z = 320 000 modulo p.
Using the same solution, we can perform these calculations with the prime having 77 digits:
p = 28 948 022 309 329 048 855 892 746 252 171 976 963 317 496 166 410 141 009 864 396 001 978 282 410 063 (such as gcd(5, p-1) = 1);
we will obtain, respectively (and almost instantly)
C = 3 355 443 200 000 000 000 000 000 000
d = 11 579 208 923 731 619 542 357 098 500 868 790 785 326 998 466 564 056 403 945 758 400 791 312 964 025,
the resulting z is indeed 320 000 modulo p.
-Remark: On the need to exclude primes congruent to 1 modulo 4.
It should be noted that we have chosen all the previous values of p to be equal to 3 (mod 4), which ensures that p is not the sum of two squares and that there are no roots of -1 modulo p; the polynomial 2x ⁴ thus becomes completely irreducible in Z_p (no extraneous factorisation or complex roots in the modulus), unlike the case where p is congruent to 1 modulo 4, where we know (Euler–Fermat’s two-squares theorem) that p is the unique sum of two squares. We would then have, on the one hand, four valid solutions (x, -x, ix, -ix) associated with the same C and leading to the same z; on the other hand, we could write, modulo p, 2x⁴ ≡ (x² + ix²)(x² - ix²), this reducibility weakening the protection of our ‘secret’. Therefore, to avoid these problems, we shall exclude primes of the form 4k + 1.
-Modular computation scheme

Given n and selecting an initial p (sufficiently large, congruent to 3 modulo 4 and such as gcd(n, p-1) = 1), the process begins with the secret number x, which is converted into a public identifier C (encrypted message); then, using the encryption key d, the resulting z is derived. Let’s summarize the three elements of the proposed implementation. - The theoretical structure of solutions of the form x = An2k(n), - The encryption function f(x) ≡ 2x4 (mod p), - The encryption mechanism, derived from Fermat's Little Theorem, z ≡ Cd (mod p). Problem: Given n, p and C (public data), find the unique value of z modulo p. We will have, successively, (x, p) → C ≡ 2x4 (mod p), (n, p) → d ≡ 1/n (mod p-1), (C, d) → z ≡ Cd (mod p).

For an observer who knows only n, p, and C, finding z without knowing the factorization of p - 1 would be practically impossible (how would one calculate d ?), whereas the designer, who chose and constructed p “on demand” and knows the factorization of p - 1, can easily obtain d and thus z.
We could have found x given z, but extracting the fourth modular root
  x ≡ (zⁿ/2)^1/4 (mod p)
is more delicate and would not allow for uniqueness as mentioned above.
We could also define a symmetric Diffie-Hellman-type protocol for our equations:
- Alice chooses a secret x, calculates C ≡ 2x⁴ (mod p), and sends it to Bob.
- Bob chooses a secret z, calculates C' ≡ zⁿ (mod p) and sends it to Alice.
If Alice and Bob manage to agree on a configuration where C = C', without revealing their secrets (x and z, respectively), they will have created a secure communication channel of the shared secret type.

- CONCLUSION
For any odd number n greater than 1, there always exist, at least for x = y, an infinite number of non-trivial solutions in Z; for even numbers n, there are no non-trivial solutions.
These solutions, that we have explained, which are most often proportional to powers of 2, are indeed infinite in number, but they are not primitive solutions since gcd(x, y, z) > 1, which is consistent with Beal’s conjecture ([11]).
The exceptional and most interesting case, n = 5, also has solutions where x and y are not proportional; this is the only known case with an infinite number of non-proportional solutions.
For the case x = y with odd exponents n greater than 1, we have shown that all non-trivial solutions are necessarily of the general form x = y = Aⁿ2^k(n), z = A⁴2^m(n), where we have specified the various elements.
By transposing these equations from Z into the finite field Z/pZ, where p is a prime number (to be chosen large, congruent to 3 modulo 4 and such that gcd(n, p-1) = 1), we propose an application to asymmetric cryptography by constructing a one-way function and describe the modular computation scheme that allows us, given n, p and C ≡ 2x⁴ (mod p), the encrypted message from the secret x, to obtain the unique solution z modulo p.

---

(+) Proof of z ≡ C^d (mod p).
Recall that Fermat’s Little Theorem states that if p is a prime number and z is an integer that is not a multiple of p, then
  z^p-1 ≡ 1 (mod p).
Since gcd(n, p-1) = 1, there exists an integer d (the inverse of n) such that
  nd = 1 + k(p - 1), where k is any integer,
which can also be written as
  nd ≡ 1 (mod p-1), or d ≡ 1/n (mod p-1).
We can raise zⁿ ≡ C (mod p) to the power of d
  (zⁿ)^d ≡ C^d (mod p),
hence
  z^1+k(p-1) = z (z^p-1)^k ≡ C^d (mod p),
and by Fermat’s Little Theorem, we indeed have
  z ≡ C^d (mod p).

---

(*)Honorary professor Paul Sabatier University, Toulouse, France, Webmaster of the site SAYRAC , E-mail: renelouis.clerc@free.fr.
A version of this text was published on zenodo.org on 3 june 2026, https://zenodo.org/records/20529889 .

---

- REFERENCES
[1] L. J. Mordell, On the rational solutions of the indeterminate equations of the third and fourth degrees, Proc. Cam. Phil. Soc., vol. 21, p. 179-192, 1922.
[2] K. Gyory, Uber die diophantische Gleichung x^p+y^p=cz^p, Publ. Math. Debrecen, 13, 301-306, 1966.
[3] K. Gyory, On the diophantine equation , Mat. Lapok, 18, 93-96, 1967.
[4] G. Kümmer, Œuvres complètes, Tome 1, éditées par A. Weil, Springer Verlag, 1975.
[5] B. J. Powell, Sur l’´equation diophantienne x^4+-y^4=z^p , Bull. Sc. Math., 107, 219-223, 1983.
[6] B. J. Powell, Proof of the Impossibility of the Fermat Equation X^p + Y^p = Z^p for Special Values of p and of the More General Equation bX^n +cY^n = cE^n, J. Number Theory 18, pp. 34-40, 1984.
[7] H. Darmon, The equation x^4-y^4=cz^p, C.R. Math. Rep. Acad. Sci. Canada. XV, No. 6, 286-290, 1993.
[8] H . Darmon, A. Granville, Sur les équations z m = f(x, y) et Ax^p + By^q = Cz^r, Bull. London Math. Soc. , 27 , pp. 513 - 543, 1995.
[9] A. Wiles, Modular elliptic curves and Fermat's last theorem, Ann. Math., Princeton University, vol. 141,‎ p. 443-551 (DOI 10.2307/2118559), 1995.
[10] H. Darmon, L. Merel, Winding quotients and some variants of Fermat's Last Theorem, Journal für die reine angew. Math. 490, 1997.
[11] R. Daniel Mauldin. A Generalization of Fermat’s Last Theorem: The Beal Conjecture et Prize Problem. Notice of AMS, Vol 44, n◦11, pp 1436-1437, 1997.
[12] K. Gyory, A. Petho and V. T. Sos, Number Theory, Diophantine, Computational and Algebraic Aspects, Walter de Gruyter, Berlin-New York, 1998.
[13] D.Hilbert, The theory of algebraic number fields, Springer-Verlag,1998.
[14] F. Beukers, The Diophantine equations x^p+y^p=z^q, Duke Mathematical Journal, Vol. 91, No. 1, 1998.
[15] N. Bruin, On powers as sums of two cubes, in Algorithmic Number Theory, ANTS-IV, springer, 2000.
[16] B. G. Sloss, A Note on a Diophantine Equation Considered by Powell, The Fibonaci quaterly, vol. 40 (3), pp. 255-258, 2002.
[17] L. V. Dieulefait, Modular congruences, Q-curves, and the Diophantine equation x^4+-y^4=z^p, Bull. Belg. Math. Soc. Simon Stevin, 12(3), pp. 363–369, 2005.
[18] Diana Savin, About the diophantine equation x^4-q^4 = py^r, arXiv:0907.077[mathN.T.], 2009.