Polynomial solutions to $A^4+B^4=C^4+D^2\,$ leading to numerical solutions to $w^4+x^4 = y^4+z^4$?
问题内容
An interesting MSE post was recently made by Koushik Pramanik. To give some background, there seems to be only one known polynomial solution to the equation in the first part of the title, namely,
$$(17 p^2 - 12 p q - 13 q^2)^4 + (17 p^2 + 12 p q - 13 q^2)^4 = (17 p^2 - q^2)^4 + (289 p^4 + 14 p^2 q^2 - 239 q^4)^2$$
However, by carefully choosing $(p,q)$, one can solve $d^2 = 289 p^4 + 14 p^2 q^2 - 239 q^4$. For example, let $(p,q)=(33,17)$ or $(p,q)=(11,3)$, then one gets the $(4,2,2)$,
$$59^4 + 158^4 = 133^4 + 134^4$$ $$193^4 + 292^4 = 257^4 + 256^4$$
after removing common factors $8\times17$ and $8$, respectively. Hence this $d^2 = \text{quartic}$ is birationally equivalent to an elliptic curve.
Question: Can we find more polynomial solutions to $A^4+B^4=C^4+D^2$ where $(A,B,C)$ are quadratic forms?
I. The System
Koushik wanted to solve the system of quadrics,
$$c x^2 + y^2 = c z^2 - w^2$$ $$x^2 + c y^2 = c w^2 - z^2$$
Eliminating the parameter $c$, one gets the well-known quartic surface,
$$w^4+x^4=y^4+z^4$$
Since there are infinitely many solutions to this equation, therefore there are also infinitely many $c$. We choose the smallest solution to the $(4,2,2)$ with this particular order $(w,x,y,z) = (59, 158, 134, 133)$ to get $c = -\dfrac{221}{75}$. After a lot of algebra, we find a second quadratic form solution,
$$\qquad \qquad w^4+x^4 = y^4+\left(\sqrt{x^2+\frac{w^2+y^2}c}\,\right)^4$$
where $c = -\dfrac{221}{75}$ and,
$$w = 22860131504 + 3091112n + 59n^2$$ $$x = 2(42137968 + 1170088n + 79n^2$$ $$y = 2(10181471152 + 1379656n + 67n^2$$ $$z^2 = x^2+\frac{w^2+y^2}c=\text{quartic}$$
So we have $z^2 = \text{quartic}\,$ just like in the first example. The first has a square leading coefficient $289=17^2$. This one also has and is the familiar square $133^2$ from the original solution, namely,
$$z^2=-318059084780255938304 - 85703480584476928n - 3090906687648n^2 + 364751216n^3 + 133^2n^4$$
which makes it easy to solve using the tangent method. Initial solutions are,
$$n = 19684,\, -27004/5,\, -343036/29,\, -107007286912612/5524044743$$
and an infinite more. (The first three just yield the original equality after removing common factors, but the 4th yields a new solution with $15$ digits.)
II. Question
To repeat, can we find more polynomial solutions to $A^4+B^4=C^4+D^2$ where $(A,B,C)$ are quadratic forms but with the preference that $z^2=\text{quartic}$, unlike the one above, involves only small coefficients?
P.S. For those who want to data-mine Wroblewski's $(4,2,2)$ database for "smallish" $c$, this is the link.
回答 (2)
(this is not an answer to the second degree question, but we can easily get 3) solutions similar to the first one with coefficients $j,I,k<4110$ have not been found for form $jp^2+ipq+kq^2$.
My answer will partially use a more powerful family with a minimum cubic degree that can be easily transformed into smaller families of the second degree, including your example of a family located inside this family. We want to find a large parametric solution to the Diophantine equation: $$A^4 + B^4 = C^4 + D^2$$ $$A = X, \quad B = bX + a, \quad C = bX + c, \quad D = Y$$ $$\small Y^2 = X^4 + 4 a b^{3} - 4 b^{3} c X^3 + 6 a^{2} b^{2} - 6 b^{2} c^{2} X^2 + 4 a^{3} b - 4 b c^{3} X + a^{4} - c^{4}$$ After the chirngause transformation: $$Y^2 = t^4 + B_1 t^2 + B_2 t + B_3, \quad Y=t^2+\frac{B_1}{2}$$ $$t=(\frac{B_1}{2})^2\frac{1}{B_2}-B_3$$ $$\small A = 4 a^{3} b^{12} - 12 a^{3} b^{8} + 9 a^{3} b^{4} - a^{3} - 12 a^{2} b^{12} c + 12 a^{2} b^{8} c + 9 a^{2} b^{4} c - a^{2} c + 12 a b^{12} c^{2} + 12 a b^{8} c^{2} - 9 a b^{4} c^{2} - a c^{2} - 4 b^{12} c^{3} - 12 b^{8} c^{3} - 9 b^{4} c^{3} - c^{3}$$ $$\small B = b \left(4 a^{3} b^{12} - 4 a^{3} b^{8} - 3 a^{3} b^{4} + 3 a^{3} - 12 a^{2} b^{12} c - 4 a^{2} b^{8} c + 9 a^{2} b^{4} c + 3 a^{2} c + 12 a b^{12} c^{2} + 20 a b^{8} c^{2} + 3 a b^{4} c^{2} + 3 a c^{2} - 4 b^{12} c^{3} - 12 b^{8} c^{3} - 9 b^{4} c^{3} - c^{3}\right)$$ $$\small C = b \left(4 a^{3} b^{12} - 12 a^{3} b^{8} + 9 a^{3} b^{4} - a^{3} - 12 a^{2} b^{12} c + 20 a^{2} b^{8} c - 3 a^{2} b^{4} c + 3 a^{2} c + 12 a b^{12} c^{2} - 4 a b^{8} c^{2} - 9 a b^{4} c^{2} + 3 a c^{2} - 4 b^{12} c^{3} - 4 b^{8} c^{3} + 3 b^{4} c^{3} + 3 c^{3}\right)$$ $$\small D = - 48 b^{4} \left(a^{2} b^{4} - a^{2} - 2 a b^{4} c + b^{4} c^{2} + c^{2}\right) \left(2 a^{2} b^{8} - 3 a^{2} b^{4} + a^{2} - 4 a b^{8} c + a c + 2 b^{8} c^{2} + 3 b^{4} c^{2} + c^{2}\right)^{2} + \left(- 12 a^{3} b^{12} + 24 a^{3} b^{8} - 13 a^{3} b^{4} + a^{3} + 36 a^{2} b^{12} c - 24 a^{2} b^{8} c - 9 a^{2} b^{4} c + a^{2} c - 36 a b^{12} c^{2} - 24 a b^{8} c^{2} + 9 a b^{4} c^{2} + a c^{2} + 12 b^{12} c^{3} + 24 b^{8} c^{3} + 13 b^{4} c^{3} + c^{3}\right)^{2}$$
- Example $a=2, b=-1, c=-4$ $$17^4+8^4=1^4+296^2$$
- the existence of ideal families with squares is very possible, even the smallest example: $a=-11,b=-1 c=-7$ $$292^4+193^4=256^4+{\color{green}{66049}}^2$$ More solutions in the family
a b c | A B C D
-----------------------------------------------------------------
-52 -1 -26 | -1 1 1 1
-51 -2 -45 | -8 -1 1 64
-45 -2 -39 | -13 -19 -13 361
-45 -1 -28 | -3364 4849 4288 18515809
-45 -1 -17 | -4849 3364 4288 18515809
-44 -1 -28 | -193 292 256 66049
-44 -1 -16 | -292 193 256 66049
-41 -3 -40 | -27 -1 1 729
-33 -1 -25 | -59 158 134 17689
-33 -1 -8 | -158 59 134 17689
-32 -1 -33 | 157 227 239 49
-32 -1 1 | 227 157 -239 49
-----------------------------------------------------------------
Total unique perfect square solutions found: 12
Codes with which to check calculations and examples
import math
def get_primitive_sol(a, b, c):
b4 = b**4
b8 = b**8
b12 = b**12
A = (4*a**3 * b12 - 12*a**3 * b8 + 9*a**3 * b4 - a**3 -
12*a**2 * b12 * c + 12*a**2 * b8 * c + 9*a**2 * b4 * c - a**2 * c +
12*a * b12 * c**2 + 12*a * b8 * c**2 - 9*a * b4 * c**2 - a * c**2 -
4*b12 * c**3 - 12*b8 * c**3 - 9*b4 * c**3 - c**3)
B = b * (4*a**3 * b12 - 4*a**3 * b8 - 3*a**3 * b4 + 3*a**3 -
12*a**2 * b12 * c - 4*a**2 * b8 * c + 9*a**2 * b4 * c + 3*a**2 * c +
12*a * b12 * c**2 + 20*a * b8 * c**2 + 3*a * b4 * c**2 + 3*a * c**2 -
4*b12 * c**3 - 12*b8 * c**3 - 9*b4 * c**3 - c**3)
C = b * (4*a**3 * b12 - 12*a**3 * b8 + 9*a**3 * b4 - a**3 -
12*a**2 * b12 * c + 20*a**2 * b8 * c - 3*a**2 * b4 * c + 3*a**2 * c +
12*a * b12 * c**2 - 4*a * b8 * c**2 - 9*a * b4 * c**2 + 3*a * c**2 -
4*b12 * c**3 - 4*b8 * c**3 + 3*b4 * c**3 + 3*c**3)
p1 = -48 * b4 * (a**2 * b4 - a**2 - 2*a * b4 * c + b4 * c**2 + c**2) * \
(2*a**2 * b8 - 3*a**2 * b4 + a**2 - 4*a * b8 * c + a*c + 2*b8 * c**2 + 3*b4 * c**2 + c**2)**2
p2 = (-12*a**3 * b12 + 24*a**3 * b8 - 13*a**3 * b4 + a**3 +
36*a**2 * b12 * c - 24*a**2 * b8 * c - 9*a**2 * b4 * c + a**2 * c -
36*a * b12 * c**2 - 24*a * b8 * c**2 + 9*a * b4 * c**2 + a * c**2 +
12*b12 * c**3 + 24*b8 * c**3 + 13*b4 * c**3 + c**3)**2
D = p1 + p2
if A == 0 or B == 0 or C == 0 or D == 0:
return None
g = math.gcd(math.gcd(abs(A), abs(B)), abs(C))
A_prim = A // g
B_prim = B // g
C_prim = C // g
D_prim = D // (g**2)
if A_prim**4 + B_prim**4 != C_prim**4 + D_prim**2:
return None
return (A_prim, B_prim, C_prim, D_prim)
def is_perfect_square(n):
if n < 0:
return False
root = math.isqrt(n)
return root * root == n
def search_solutions(limit=2):
unique_solutions = set()
print(f"{'a':>3} {'b':>3} {'c':>3} | {'A':>10} {'B':>10} {'C':>10} {'D':>12}")
print("-" * 65)
for a in range(-limit, limit + 1):
for b in range(-limit, limit + 1):
if b == 0:
continue
for c in range(-limit, limit + 1):
if a == c:
continue
sol = get_primitive_sol(a, b, c)
if sol is None:
continue
key = (abs(sol[0]), abs(sol[1]), abs(sol[2]), abs(sol[3]))
if key not in unique_solutions:
unique_solutions.add(key)
output_line = f"{a:>3} {b:>3} {c:>3} | {sol[0]:>10} {sol[1]:>10} {sol[2]:>10} {sol[3]:>12}"
if is_perfect_square(sol[3]):
print(f"\033[92m{output_line}\033[0m")
else:
print(output_line)
print("-" * 65)
print(f"Total unique solutions found: {len(unique_solutions)}")
if __name__ == "__main__":
search_solutions(limit=2)
import sympy as sp
def generate_latex_solution():
a, b, c = sp.symbols('a b c', rational=True)
X, Y, t = sp.symbols('X Y t')
rhs_x = sp.expand(X**4 + (b*X + a)**4 - (b*X + c)**4)
j = rhs_x.coeff(X, 3)
shift = j / 4
rhs_t = sp.expand(rhs_x.subs(X, t - shift))
b1 = rhs_t.coeff(t, 2)
b2 = rhs_t.coeff(t, 1)
b3 = rhs_t.coeff(t, 0)
t_sol = sp.simplify(((b1 / 2)**2 - b3) / b2)
x_sol = sp.simplify((t - shift).subs(t, t_sol))
A_sol = x_sol
B_sol = sp.simplify(b * x_sol + a)
C_sol = sp.simplify(b * x_sol + c)
D_sol = sp.simplify(t_sol**2 + b1 / 2)
diff = sp.simplify((A_sol**4 + B_sol**4) - (C_sol**4 + D_sol**2))
assert diff == 0, "Validation failed."
den_A = A_sol.as_numer_denom()[1]
den_B = B_sol.as_numer_denom()[1]
den_C = C_sol.as_numer_denom()[1]
L = sp.lcm(sp.lcm(den_A, den_B), den_C)
A_poly = sp.simplify(A_sol * L)
B_poly = sp.simplify(B_sol * L)
C_poly = sp.simplify(C_sol * L)
D_poly = sp.simplify(D_sol * (L**2))
mse_post = f"""
We want to find a parametric solution to the Diophantine equation:
$$A^4 + B^4 = C^4 + D^2$$
Using the clever substitution suggested by the Tschirnhausen transformation:
$$A = X, \\quad B = bX + a, \\quad C = bX + c, \\quad D = Y$$
Substituting these into the equation and expanding gives a quartic equation where the $X^4$ coefficients nicely reduce to $1$:
$$Y^2 = X^4 + {sp.latex(j)} X^3 + {sp.latex(rhs_x.coeff(X, 2))} X^2 + {sp.latex(rhs_x.coeff(X, 1))} X + {sp.latex(rhs_x.coeff(X, 0))}$$
Applying the shift $X = t - {sp.latex(shift)}$ eliminates the $t^3$ term, yielding:
$$Y^2 = t^4 + B_1 t^2 + B_2 t + B_3$$
By substituting $Y = t^2 + \\frac{{B_1}}{{2}}$, the higher-order terms $t^4$ and $t^2$ cancel completely, leaving a linear equation in $t$. Solving this linear equation, back-substituting, and multiplying through by the common denominator to clear fractions, we obtain the following **purely polynomial integer solution** in terms of free parameters $a, b, c$:
$$A = {sp.latex(A_poly)}$$
$$B = {sp.latex(B_poly)}$$
$$C = {sp.latex(C_poly)}$$
$$D = {sp.latex(D_poly)}$$
Using algebraic identity, it can be verified that $A^4 + B^4 - C^4 = D^2$ holds universally for any $a, b, c$.
"""
print(mse_post.strip())
if __name__ == "__main__":
generate_latex_solution()
```
$$A^4+B^4=C^4+D^2 \tag{1}$$
Let $A=x+a, B=x+b, C=x+c, D=x^2+qx+r.$
We consider the equation $(1)$ as a quadratic equation in $x.$
To eliminate the coefficient of the $x^3$ and $x^2$ term, we determine $q,r,$ and we get $x,$
\begin{align*} q &=-2c+2b+2a,\\ r &=-5c^2+4cb+4ca+b^2-4ba+a^2,\\ x &= -\dfrac{4a^2-9ba+7ca+4b^2+7cb-13c^2}{6(a+b-2c)}. \end{align*}
This allows us to derive the parametric solution of degree four.
\begin{align*} A &=2a^2+(-19c+15b)a-7cb+13c^2-4b^2,\\ B &=-4a^2+(-7c+15b)a-19cb+13c^2+2b^2,\\ C &=-4a^2+(-c+9b)a-cb+c^2-4b^2,\\ D &=4a^4+(116c-132b)a^3+(17b^2-355c^2+362cb)a^2 \\ &+(-132b^3+478c^3+362cb^2-724c^2b)a \\ &+4b^4-239c^4+116cb^3+478c^3b-355c^2b^2. \end{align*}
Numerical solutions where $(b,c)<50$ and height$(a)<1000.$
a b c A B C D
[15/4, 1, 2] [-193, -292, -256, 257]
[-31, 1, 2] [157, -227, -239, 7]
[41/8, 1, 2] [-59, -158, -134, 133]
[62/17, 1, 2] [-3364, -4849, -4288, 4303]
[158/13, 1, 2] [248, -2797, -2524, 2131]
[199/33, 1, 2] [-34813, -134413, -114613, 111637]
[415/43, 1, 2] [-3119, -641471, -567683, 505829]
[-707/188, 9, 8] [100019, -756424, -689308, 564749]