1is it possible???
i am not getting ny such integer satisfying this equation!!...
1We will get a third degree eqn in N.. → N^3- 6N^2- 2=0.
If we consider roots of this eqn a,b,c…
a+b+c=6 ---(1),
abc=2 ---(2),
ab+bc+ca=0 ---(3).
From these we get, a^2 +b^2 +c^2=36 ----(4).
No “integer†solutions satisfy all these….so no solution…
62This is a good way to solve...
But, I dont know how this directly implies the result.. i guess u just din write a couple of steps that to you were obvious?
62btw i think there is a better solution:
N3- 6N2- 2=0
N3 = 6N2 + 2
So when N is odd, LHS is odd and RHS is even .. they cant be equal.
N is even so substitute N=2k
8k3 = 6.4 k2 + 2
dividing by 2,
4k3 = 6.2 k2 + 1
LHS is even... RHS is odd..
Hence, no solution exists!
1haan well nice one...
actuallly wat was my interpretation...
for eqn(2) to satisfy,,, a,b,c ε(-1,2)
agn two of them cant be 2 at a time.
watever it be,,, 12+12+22 will never be equal to 36 as in eqn(4)...
so no integer possible..
nyway it is rather logical..
urs is a finer one...
62Good work...
a couple of small things.. but i think u just lazed to write them down :)
