a = x^3/2 b= y^3/2 => (a-b)(a+b) = 13*7 => a=10, b=3 => x^3 = 100, y^3 = 9 => if cube root of unity are 1 w w^2 then x = 100^1/3, w100^1/3, (w^2)100^1/3 and y = 9^1/3, w9^1/3, (w^2)9^1/3.
By observation, X and Y cannot be fractional as the subtraction of cubes is not fractionsl! So the X and Y have to be whole numbers. Also either X or Y have to have cube more than 91! So the first number to satisfy this is 5 as cube 5 is 125! If X is 5 then cube of Y must be 125+/- 91 i.e. 34 or 216! Obviously 216 is cube of 6 and hence the answer is x=6 and y =5!
(x-y)(x²+xy+y²)=1*91 is not necessary equivalent to x-y=1 and x²+xy+y²=91. It is not rigorous. If you do that, then you have to prove that there are no more solutions. Besides, you see for yourself that I imitate you and have: (x-y)(x²+xy+y²)=91=7*13 => x-y=7,x²+xy+y²=13. This gives 2 solutions (x; y)=(3; -4), (4; -3). If x, y are integers, then of course it is enough to consider the system of equations: x-y=-91, -13, -7, -1, 1, 7, 13, 91 x²+xy+y²=-1, -7, -13, -91, 91, 13, 7, 1 In this task, x, y are real. If we use a nonequivalent transformation, then we have to prove that there are no more solutions.
Interesting question
Excellent method
Nice method and you also verify brilliant
Think outside the box: 27+64=91.
a = x^3/2 b= y^3/2 => (a-b)(a+b) = 13*7 => a=10, b=3 => x^3 = 100, y^3 = 9 => if cube root of unity are 1 w w^2 then x = 100^1/3, w100^1/3, (w^2)100^1/3 and y = 9^1/3, w9^1/3, (w^2)9^1/3.
By observation, X and Y cannot be fractional as the subtraction of cubes is not fractionsl! So the X and Y have to be whole numbers. Also either X or Y have to have cube more than 91! So the first number to satisfy this is 5 as cube 5 is 125! If X is 5 then cube of Y must be 125+/- 91 i.e. 34 or 216! Obviously 216 is cube of 6 and hence the answer is x=6 and y =5!
(x-y)(x²+xy+y²)=1*91 is not necessary equivalent to x-y=1 and x²+xy+y²=91. It is not rigorous.
If you do that, then you have to prove that there are no more solutions. Besides, you see for yourself that I imitate you and have:
(x-y)(x²+xy+y²)=91=7*13 => x-y=7,x²+xy+y²=13. This gives 2 solutions (x; y)=(3; -4), (4; -3).
If x, y are integers, then of course it is enough to consider the system of equations:
x-y=-91, -13, -7, -1, 1, 7, 13, 91
x²+xy+y²=-1, -7, -13, -91, 91, 13, 7, 1
In this task, x, y are real. If we use a nonequivalent transformation, then we have to prove that there are no more solutions.
i agree, there are infinite solutions for x and y