Diophantine equationsI 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 =...

Diophantine equations

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

Thank you!

Asked on by fryvonne

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oldnick | (Level 1) Valedictorian

Posted on

`3)x^3-y^3=91`   (1)

we can wrtite it as:

`x^3-y^3=(x-y)(x^2+xy+y^2)`

We are awating that both  `(x-y) `  than `(x^2+xy+y^2)`  dividing 91.

So the couple they have to assume are:

`(1,91),(7,13),(13,7),(91,1)`

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.

So: 

`a)`   if   `x-y=1`   `y=x-1`

substituing in the second member:

`x^2 +x(x-1)+(x-1)^2= 91`

`x^2+x^2-x+x^2-2x+1=91`

`3x^2-3x+1=91`

`3x^2-3x-90=0`

`x^2-x-30=0`

`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`

substituing:       `x^2+x(x-7)+(x-7)^2=13`

`x^2+x^2-7x+x^2-14x+49=13`

`3x^2-21x+39=0`

`x^2-7x+13=0`

`Delta=49-4(13)=-3<0`  

hasn't real soluions, so neither numeric one.

`c)`      `x-y=13`    `y=x-13`

substituing:      `x^2+x(x-13)+(x-13)^2=7`

`x^2+x^2-13x+x^2-26x+169=7`

`3x^2-39x+162=0`

`x^2-13x+108=0`

`Delta=169-4(108)=-263<0`

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)`

 

 

 

 

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oldnick | (Level 1) Valedictorian

Posted on

`2)`    `x+y=xy`

We can write this equation as:

`xy-x-y=0`

developing:

`(x-1)(y-1)-1=0`

`(x-1)(y-1)=1`

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.

oldnick's profile pic

oldnick | (Level 1) Valedictorian

Posted on

`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`

`4x-6y+11z=1```

So if we set     `x_0=-7,y_0=7,z_0=7`

is one solution of equation indeed:

indeed:

`4x_0-6y_0+11z_0=7`

By Chinese's Rimander Theorem, we can get other solution:

`x'_0=-7+11t_1`

`y'_0=7+11t_2`

`z'_0= 7-4t_1+6t_2`

where `t_1,t_2,t_3`  are arbitray integers.

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

x^3-Y^3=91.

(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+1)^2+y^2+(y+1)y-91=0

3y^2+3y-90=0

y^2+y-30=0

(y+6)(y-5)=0

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.

 

 

 

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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 systems

x - 1 = 1, y - 1 = 1, x - 1 = -1, y - 1 = -1,

solutions being (0,0) si (2,2).

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