Question

\[x^3-y^3=xy+61\]
\[x,y \in \mathbb{N}\]

Answer 1

what restrictions are there for \(x\) and \(y\) ?

Answer 2

they both should be natural numbers :P

Answer 3

lol

Answer 4

haha nice !!

Answer 5

and they should obviously satisfy the eqn

Answer 6

nice man :D

Answer 7

i mean \(x-y\) must be a limited natural number ... why?

Answer 8

whay x can not be very greater than y ?

Answer 9

i dont get it

Answer 10

if x>>y then LHS will be very greater than RHS due exponents... 3(LHS) in comparison with 2(RHS)

Answer 11

*due to exponents

Answer 12

so let\[x-y=a\]a must be a restricted number equation becomes to :\[y^2(3a-1)+y(3a^2-2)+a^3-61=0\]

Answer 13

y is positive and if a>something then LHS of later equation >0 and there is no solution so whats that something now?

Answer 14

x and y both can be any numbers as long as they satisfy the equation

Answer 15

we want to find all possible values of x and y that satisfies the equation

Answer 16

if a>4 there is no solution for the later equation am i right?

Answer 17

sorry man i am not getting it :(

Answer 18

np man

Answer 19

even we can tell if a>3 there is no solution for the later equation

Answer 20

so \[a=1,2,3\]just from \(a=1\) we get the solution \((x,y)=(6,5)\).

Answer 21

and there is no solution from \(a=2,3\)

Answer 22

how did u get y(3a^2 - 2) ??
i m getting y(3a^2-a).....

Answer 23

sorry thats right\[y^2(3a-1)+y(3a^2-a)+a^3-61=0\]

Answer 24

a>3 no solution.....does it have something to do with discriminant of that equation?how did u get a>3 ,no solution?

Answer 25

if a>3 all the coefficients of quadratic will be positive ... y is a positive number too...so the LHS became a positive expression

Answer 26

and of course we workin with natural numbers here

Answer 27

yes,i got that....thanks :)

Answer 28

:)