# 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 =...

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) 4*x* - 6*y* + 11*z* = 7

2) x + y = xy

3) y3 - x3 = 91

Thank you!

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`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)`

`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.

`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.

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.

1) 4*x* - 6*y* + 11*z* = 7

We'll divide -6 by 4 =>-6 = 4(-2) + 2,

The initial equation will become

4(*x* - 2*y*) + 2*y* + 11*z* = 7.*We'll substitute x*' = *x* - 2*y* and we'll obtain 4*x'*+ 2*y* + 11*z* = 7.

But 11 = 2·5 + 1, we'll write the last version of the equation

4*x'* + 2(*y* + 5*z*) + *z* = 7.

We'll substitute *y'* = *y* + 5*z* from the above equation

4*x'* + 2*y*' + *z* = 7.

Solutions of this equation are *z* = 7 - 4*x*'- 2*y'*, *x*'si *y'* arbitrary whole numbers.

So *y* = *y'* - 5*z* = 20*x'* + 11*y*' - 35,

*x* = *x*' + 2*y* = 41*x*' + 22*y*' - 70.

The solution of the initial equation has the form

*x* = 41*x'* + 22*y*' - 70 *y* = 20*x*' + 11*y*'- 35 *z* = 7 - 4*x*' - 2*y'*,

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).