2020/022) Find integers $x,y$ such that $x^3-y^3=91$

We have $x^-y-y^3 = 91$

or $(x-y) (x^2+xy+y^2) = 91= 13 * 7$

clearly $x >=y$

So factor of 91 =  1 * 91 and 7 * 13

 so we have following 4 cases

case 1:

$x-y= 1\cdots(1)$

and $x^2 + xy + y^2 =  = 91\cdots(2)$

From (1) we have $x=y+1$

putting in (2) we get $(y+1)^2 + y(y+1) + y^2 = 91$

Or $3y^2 + 3y = 90$

or $y^2+y-30=0$

or $y^2+y-30=0$

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

so y = 5 or -6 and x = y+ 1 gives 2 solutions

$(6,5)$ and $(-5,-6)$

case 2:

$x-y= 7\cdots(1)$

and $x^2 + xy + y^2 =  = 13\cdots(2)$

From (1) we have $x=y+7$

putting in (2) we get $(y+7)^2 + y(y+7) + y^2 = 13$

Or $3y^2 + 21y + 49 = 13$

or $3y^2+21y+36=0$

or $y^2+7y+12 =0$

or$(y+3)(y+4)=0$

so y = -3 or -4  and x = y+ 7 gives 2 solutions

$(4,-3)$ and $(3,-4)$

 

case 3:

$x-y= 13\cdots(1)$

and $x^2 + xy + y^2 =  = 7\cdots(2)$

From (1) we have $x=y+13$

putting in (2) we get $(y+13)^2 + y(y+13) + y^2 = 7$

Or $3y^2 + 39y + 169 = 7$

or $3y^2+39y+162=0$

or $y^2+13y+54 =0$

This does not have integer solution

case 4:

$x-y= 91 \cdots(1)$

and $x^2 + xy + y^2 =  = 1\cdots(2)$

From (1) we have $x=y+91$

putting in (2) we get $(y+91)^2 + y(y+91) + y^2 = 1$

Or $3y^2 + 273y + 8281 = 1$

or $3y^2+ 273y+ 8280=0$

or $y^2+91y+ 2760 =0$

ad $91^2-8 * 2760 < 0$ this does not have any real solution

This does not have integer solution

So solution sets are  $(6,5),(-5,-6),(4,-3),(3,-4)$

Short cut solution:(can be applied for objective question)

we need to find the limit of x and y

let is take the difference of $(x+1)^3 and $x^3$ this keeps on increasing when x increases for positive x

and when decreases for -ve x .

so we can bound x between -5 and 6 as $7^3-6^3 > 91$

by putting the value of x from -5 to 6 we can find the value of y as well and solution pair as above