If we have the equation x^3+y^3+z^3=2,will it be possible to have integer solutions?
A first look will make you think that there are none,but there are infinite number of solutions
You put x=1+p and y=1-p.You can then work out z as well as the value of p.
x^3+y^3=(1+3p+3p^2+p^3)+(1-3p+3p^2-p^3)=2+6p^2.
So we must have z^3=(-)6 p^2.
If p=(-)6a^3 ,we can get z^3=(-)6*(36a^6)=(-)216 a^6 and so z=(-)6 a^2
Thus the solutions are x=1-6a^3,y=1+6a^3 and z=(-)6 a^2 for any value 'a' can take
Example if a=2,we have x=1-48=(-)47,y=1+48=49 and z=(-)24
x^3+y^3+z^3=(49^3)-(47^3)-(24^3)=117649-103823-13824=2.
A first look will make you think that there are none,but there are infinite number of solutions
You put x=1+p and y=1-p.You can then work out z as well as the value of p.
x^3+y^3=(1+3p+3p^2+p^3)+(1-3p+3p^2-p^3)=2+6p^2.
So we must have z^3=(-)6 p^2.
If p=(-)6a^3 ,we can get z^3=(-)6*(36a^6)=(-)216 a^6 and so z=(-)6 a^2
Thus the solutions are x=1-6a^3,y=1+6a^3 and z=(-)6 a^2 for any value 'a' can take
Example if a=2,we have x=1-48=(-)47,y=1+48=49 and z=(-)24
x^3+y^3+z^3=(49^3)-(47^3)-(24^3)=117649-103823-13824=2.
