这是一个线性丢番图方程

Question

我想校准

https://en.wikipedia.org/wiki/Diophantine_equation#Examples

ax+by=c这是一个线性丢番图方程。

我试着

我认为你替换h的方式不好。我该怎么修？

x、 y整数

(1( (2(5^4x-2^4y=1

(3( 5^5x-2^5y=1

(4( 11^5x-2^5y=1

from sympy.solvers.inequalities import reduce_rational_inequalities
from sympy import *
import math
var('x y n')
def myEq(f,g,h,myMax):
myN = Symbol('myN', real=True)
myGcdex = gcdex(f, -g)
myX = g * n + myGcdex[2]
myY = solve((f * x - g * y - h).subs({x: myX}), y)[0]
myN = math.ceil(reduce_rational_inequalities([[myX.subs({n: myN}) >= myMax]], myN).lhs)
return myX.subs({n: myN}),myY.subs({n: myN})
print("#(1)",myEq ( 5**4,2**4,1,  0) )
print("#(2)",myEq ( 5**4,2**4,1, 10) )
print("#(3)",myEq ( 5**5,2**5,1,100),    "????＜correct＞x=125,y=12207")
print("#(4)",myEq (11**5,2**5,1,  0)," ","????＜correct＞x= 19,y=95624")
#(1) (1, 39)
#(2) (17, 664)
#(3) (129, 100781/8) ????＜correct＞x=125,y=12207
#(4) (1, 80525/16)   ????＜correct＞x= 19,y=95624
#
#〇(1)nxy 0 16*n + 1 625*n + 39          ＜correct＞x = 16 n + 1  , y = 625 n + 39      , n=0
#〇(2)nxy 1 16*n + 1 625*n + 39
#×(3)nxy 4 32*n + 1 3125*n + 781/8      ＜correct＞x = 32 n + 29 , y = 3125 n + 2832   , n=3
#×(4)nxy 0 32*n + 1 161051*n + 80525/16 ＜correct＞x = 32 n + 19 , y = 161051 n + 95624, n=0

参考

我想要从xy到单元格的函数转换

https://mathworld.wolfram.com/Congruence.html

(20220407(

MY_diox( 5**4*x-2**4*y-1,  0) # ((  1,    39), (16*n - 15,    625*n -   586))
MY_diox( 5**4*x-2**4*y-1, 10) # (( 17,   664), (16*n +  1,    625*n +    39))
MY_diox( 5**5*x-2**5*y-1,100) # ((125, 12207), (32*n + 93,   3125*n +  9082))  
MY_diox(11**5*x-2**5*y-1,  0) # (( 19, 95624), (32*n - 13, 161051*n - 65427))

(20220410(

参考

获取线性方程的所有正积分解

Answer 1

看起来你正在试图获得方程的特定解，该方程对x具有特定的最小值。如果你求解参数的不等式，你可以将其转换为整数上的区间，只需选择第一个这样的整数。也可以使用原点偏移来给出在n = 1处给出该值的修正方程。

def diox(eq, x):
"""
>>> from sympy.abc import x, y
>>> eq = 5**5*x-2**5*y-1
>>> diox(eq, 30)
((61, 5957), (32*n + 29, 3125*n + 2832))
"""
s = diophantine(eq)
assert len(s) == 1
a, b = s.pop()
p = a.free_symbols
assert len(p) == 1
p = p.pop()
i = solve(a >= x).as_set().intersection(Integers)
if i.start.is_infinite:
i = i.sup
else:
i = i.start
t = Tuple(a, b).xreplace({p: i})
e = Tuple(a, b).xreplace({p: Symbol('n')+i-1})
return t, e

Answer 2

正如预期的那样，以下操作是不可能的。

from sympy import *
from sympy.abc import x, y
from sympy.solvers.diophantine import diophantine
var('x a b c')
myFormula= a*x+b*y-c
def diox(eq, x):
.......................................     
return t, e
print("#",diox(myFormula,0))
# NotImplementedError: No solver has been written for inhomogeneous_general_quadratic.