(x^3)-(y^3)=91
- Thread starter Tuugii
- Start date
Thank you very much Denis.
Unfortunately I'm not genius as Denis to obtain such an astute formula to hide, but sitting for a while got the following result:
Let's write: (y^3-x^3)=(y-x)(y^2+xy+x^2)=13*7=91; (1)
(y^2+xy+x^2)=((y+x/2)^2)+(3x^2)/4, which is >=0, thus it says that for all values of x and y the "(y^2+xy+x^2)" has a value>0;
If there are integer values x,y for the equation (1), then both multiples of the left hand side will get integer values. The right hand side of the equation (1) is divided by 7 and 13, which means the left hand side must also. Since the numbers 7 and 13 are prime numbers, there are only 4 following cases:
1) y-x is divided by 7 and 13. So, y-x=91 and (y^2+xy+x^2)=1. But this system doesn't have any values for real numbers x and y.
2) (y^2+xy+x^2) is divided by 7 and 13. So, y-x=1 and (y^2+xy+x^2)=91. Solving this system, we'll obtain x=5, y=6; x=-6, y=-5.
3) y-x is divided by 13 and (y^2+xy+x^2) is divided by 7. For this case y-x=13 and (y^2+xy+x^2)=7. This system is also doesn't have any real values.
4) y-x is divided by 7 and (y^2+xy+x^2) is divided by 13. So y-x=7 and (y^2+xy+x^2)=13. Solving this system we'll obtain x=-3, y=4; x=-4, y=3;
So the equation (1) has following real integer values:
x=5, y=6; x=-6, y=-5; x=3, y=-4; and x=-4, y=3.
Tuugii
Unfortunately I'm not genius as Denis to obtain such an astute formula to hide, but sitting for a while got the following result:
Let's write: (y^3-x^3)=(y-x)(y^2+xy+x^2)=13*7=91; (1)
(y^2+xy+x^2)=((y+x/2)^2)+(3x^2)/4, which is >=0, thus it says that for all values of x and y the "(y^2+xy+x^2)" has a value>0;
If there are integer values x,y for the equation (1), then both multiples of the left hand side will get integer values. The right hand side of the equation (1) is divided by 7 and 13, which means the left hand side must also. Since the numbers 7 and 13 are prime numbers, there are only 4 following cases:
1) y-x is divided by 7 and 13. So, y-x=91 and (y^2+xy+x^2)=1. But this system doesn't have any values for real numbers x and y.
2) (y^2+xy+x^2) is divided by 7 and 13. So, y-x=1 and (y^2+xy+x^2)=91. Solving this system, we'll obtain x=5, y=6; x=-6, y=-5.
3) y-x is divided by 13 and (y^2+xy+x^2) is divided by 7. For this case y-x=13 and (y^2+xy+x^2)=7. This system is also doesn't have any real values.
4) y-x is divided by 7 and (y^2+xy+x^2) is divided by 13. So y-x=7 and (y^2+xy+x^2)=13. Solving this system we'll obtain x=-3, y=4; x=-4, y=3;
So the equation (1) has following real integer values:
x=5, y=6; x=-6, y=-5; x=3, y=-4; and x=-4, y=3.
Tuugii