Question

2 equations 3variables

l+m=1-n

l^3 + m^3 = 1-n^2

Answer 1

consider m=t
where t is a parameter

Answer 2

l=1
m=0
n=0

Answer 3

it is one of the solutions (;

Answer 4

rvc i didn't get it
we put m=t
then?

Answer 5

every combo
(1,0,0)
(0,0,1)
(0,1,0)

Answer 6

there are more solutions :)

Answer 7

\[l^2+m^2-lm=1+n\] this is the third eqn

Answer 8

\[(l+m)(l^2+m^2-lm)=(1+n)(1-n)\]since\[l+m=1-n\]therefore\[l^2+m^2-lm=1+n\]

Answer 9

we can clearly see that (0,0,1) is a solution

But your equation is not satisfying this solution.

Answer 10

well i have not done mistakes in using identities

Answer 11

is (0,0,1) the answer given in ur book ?

Answer 12

i will get the answers tomorrow my proff gave this prob to me . But according to my solution i am also not getting (0,01)
but it is satisfying the equation

Answer 13

i think if \[l+m=1+n\]then it is also satisfying my eqn because then i will get eqn
\[l^2+m^2-lm=1-n\]

Answer 14

\[l+m=1+n\]\[l^3+m^3=1-n^2\]\[\cancel{(l+m)}(l^2+m^2-lm)=\cancel {(1+n)}(1-n)\]

Answer 15

the answer (0,0,1) is satisfying this eqn.
actually my conclusion is that ur first eqn i.e l+m=1-n is probably wrong

Answer 16

wait if (0,0,1) satisfies you equation
but u cut that (l+m) with (1+n)
l+m=0+0
and u can't divide by 0

Answer 17

while i was cancelling i had assumed that i don't know the actual soln & cancelled them as variable.

Answer 18

(1,-1,1)
(-1,1,1)

Answer 19

this will not satisfy the 2nd equation

Answer 20

lhs and rhs =0 for both equations

Answer 21

now wait its ok

Answer 22

do l,m,n have to only be positive integers or just integers?

Answer 23

just integers

Answer 24

i read the question wrong actually
i gotta work on a new solution
the question is correct tho

Answer 25

they have to be integers? no fractions?

Answer 26

yes only integers.

Answer 27

I'm getting stuck after factoring to this point which admittedly isn't very far lol:

\[l^3+m^3=1-n^2\]
\[(l+m)(l^2-lm+m^2)=(1-n)(1+n)\]
\[l^2-lm+m^2=1+n\]

Answer 28

there are solutions apart from the ones i mentioned?

Answer 29

yes i know 6 solutions so far

Answer 30

If l and m are both even then n is odd, otherwise n is even from just looking at the equation in mod 2.

Answer 31

yes :)

Answer 32

\(n^2 =1-(l^3+m^3)\)

l^3 and m^3 has to be a negative number
implies L+m is negative

and
n=1-(L+m)
n has to be positive

Answer 33

I made this just now to brute force check solutions like this with mods... heh... https://repl.it/B1t9

Answer 34

Well not solutions, it just tells you what n can be in mod k for every choice of l and m.

Answer 35

try to eliminate n :)

Answer 36

\[l^2-lm+m^2+l+m-2=0\]
It's an ellipse, so there are only finitely many points to check so I ask WA to do it for me cause it's so nice and it does and tells me: http://www.wolframalpha.com/input/?i=l%5E2%2Bm%5E2%2Bl%2Bm-lm-2%3D0&t=crmtb01&f=rc

Answer 37

Heh ok pretty bang-up job and not very elegant solution I'm giving here, but effectively done, maybe just have to check that it actually works... uhhh I'll stop here cause there's probably a prettier way and I don't wanna continue ruining it haha.

Answer 38

well, now we have more than six solutions....

Answer 39

but now (1,-1,1) and (-1,1,1) does not satisfy
\[l^2-lm+m^2+l+m-2=0\]

Answer 40

so the solutions reduce to 6 :D
but (1,-1,1) and (-1,1,1) are solutions when we check them by putting in the original 2 equations

Answer 41

we have very few integers to check as its an ellipse

Answer 42

we are not yet taught ellipse

Answer 43

are u taught about circle yet

Answer 44

yes :)

Answer 45

ok then same thing xD

Answer 46

x'D

Answer 47

its algebra tho

Answer 48

ok ill think about a nicer solution

Answer 49

:)

Answer 50

l+m=1-n
l^3 + m^3 = 1-n^2

(l+m)(l^2+m^2-lm)=(1+n)(1-n)

(l^2+m^2-lm)=(1+n)
l^2+m^2=1+n+lm

Answer 51

brb

Answer 52

ok :)

Answer 53

what are those variables? Integers?

Answer 54

Ah you've already mentioned it :)

Answer 55

\(n=1\) gives \(l=-m\) as a family of solutions. For \(n \neq 1\) one can eliminate \(l\) easily to get\[(m+n-1)^2+m^2=1+n\]Simplify to get a quadratic in terms of \(m\):\[2m^2+(2n-2)m+n^2-3n=0\]look at the discriminant, in order to get integer values for \(m\) it must be non-negative and integer:\[\Delta_m=-n^2+4n+1 \ge0\]It gives \(n=2, 3, 4\) (why?) only, so you a few cases to check.

Answer 56

I better be careful, last line must be: It gives \(n=0, 2, 3, 4\) (why?) only, so you have a few cases to check.

Answer 57

yes we can conclude that n lies in the interval [0,4]

Answer 58

one of the solutions kainui has shown is
l=-3
m=-2
n=6

Answer 59

ok ima tell the solution :)

Answer 60

eliminating n from the equation we get-\[l^2+m^2+[1-(l+m)]^2=1\]after factorization-\[(l+m)(l^2 -lm+m^2+l+m-2)=0\]Now

CASE 1--->
suppose l+m=0 then n=1
and (l, m, n)=(-a, +a ,1) here a is an integer this will give us some solutions

CASE 2---->
suppose l+m≠0 then we must have -\[l^2 -lm +m^2+l+m-2=0\]and we ccan write it like this- :)\[(2l-m+1)^2+3(m+1)^2=12\]now we have 2 possibilities \[2l-m+1=0, y+1=\pm2\]and\[2l-m+1=\pm3,    y+1=\pm1\]

Now analyzing all the case we get
(l, m, n)=(1,0,0), (0,1,0), (-2,-3,6), (-3, -2, 6), (0, -2,3 ), (-2, 0, 3)

Answer 61

WOW