I started a study concerning dyophantine equations. For this purpose I would want to propose some examples, in order to be solved in Diophantine manner.
1) 4x - 6y + 11z = 7
2) x + y = xy
3) y3 - x3 = 91
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we can wrtite it as:
We are awating that both `(x-y) ` than `(x^2+xy+y^2)` dividing 91.
So the couple they have to assume are:
for 1,7,13 ,91 are the lonely divisors of 91.
from the first members `(x-y)` we can suppose assumes step on step divisors values.
`a)` if `x-y=1` `y=x-1`
substituing in the second member:
`x^2 +x(x-1)+(x-1)^2= 91`
`Delta=1-4(-30)= 121>0` has two real solution.
`x=( 1+-11)/2` `x_1=6` `x_2=-5`
so the couples `(6,5)` and `(-5,-6)` are both solutions of (1)
`b)` `x-y=7` `y=x-7`
hasn't real soluions, so neither numeric one.
`c)` `x-y=13` `y=x-13`
Has'nt real solution thus neither integer solutions.
Then the only solution of equation `x^3-y^3=91` are the couples:`(6,5)` and `(-5;-6)`
We can write this equation as:
Last equation show at once number solution `x,y` has to be so that the modulus of their product is equal to 1
That means both `(x-1)` than `(y-1)` are to be of modulus =1
so: `x-1=1` `y-1=1` `rArr` `x=2` `y=2`
or: `x-1=-1` `y-1=-1` `rArr` `x=0` `y=0`
These are the lonely integer solutions.
`1)` First the eqaution havesolution if only if GCD( 4,6,11)|7
Indeedits so .
Now we have to note that if we set `x=-1,y=1,z=1`
So if we set `x_0=-7,y_0=7,z_0=7`
is one solution of equation indeed:
By Chinese's Rimander Theorem, we can get other solution:
where `t_1,t_2,t_3` are arbitray integers.
(x-y)(x^2+y^2+xy)=91 = 7*13 =1*91.
I take the choice of equating x-y=7 and x^2+y^2+xy=91 and examine whether there is any integral solution.
x-y=1. so x= y+1, Making this substititution inthe other factor, we get:
y=5 is a satisfying positive integer solution.
y=-6 is satisfying negative integer.
x= y+1 gives corresponding x values
x=6 when y=5
x=-5 when y=-6.
Try with other factors like x-y=7 and x^2+y^2+xy=13 and see what happens.
1) 4x - 6y + 11z = 7
We'll divide -6 by 4 =>-6 = 4(-2) + 2,
The initial equation will become
4(x - 2y) + 2y + 11z = 7.We'll substitute x' = x - 2y and we'll obtain 4x'+ 2y + 11z = 7.
But 11 = 2·5 + 1, we'll write the last version of the equation
4x' + 2(y + 5z) + z = 7.
We'll substitute y' = y + 5z from the above equation
4x' + 2y' + z = 7.
Solutions of this equation are z = 7 - 4x'- 2y', x'si y' arbitrary whole numbers.
So y = y' - 5z = 20x' + 11y' - 35,
x = x' + 2y = 41x' + 22y' - 70.
The solution of the initial equation has the form
x = 41x' + 22y' - 70 y = 20x' + 11y'- 35 z = 7 - 4x' - 2y',
where x', y' arbitrary whole numbers.
2) x + y = xy.
The initial eq. is written in this way(x - 1)(y - 1) = 1.
The product of two whole numbers is equal to 1 only if both numbers are equal to 1 or equal to -1. We'll obtain the systemsx - 1 = 1, y - 1 = 1, x - 1 = -1, y - 1 = -1,
solutions being (0,0) si (2,2).
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