Question

Doubt! :/
[Question attached]

Answer 1

I would try solving by
setting x = a^2 + 4ab + b^2
and y = c^2 + 4cd + d^2
multiply out, and see if you can re-write the product in the required form

Answer 2

I tried out the exact same thing. But 12abcd is always left out. :/

Answer 3

I'll think about it.

Answer 4

x=s^2+sd+d^2
y=e^2+er+r^2

Answer 5

so just find xy

Answer 6

xy=(s^2+sd+d^2) (e^2+er+r^2)

Answer 7

xy = (s^2+sd+d^2)(e^2+er+r^2) ?
I tried that 5 times now and I couldn't put in in the form od a2 + 4ab + b2. :/

Answer 8

@mathslover
@shamil98

Answer 9

\(\large a^2 + 4ab + b^2 = (a+2b)^2 - 3b^2\)

change of coordinates :
\(m = a+2b\)
\(n = b\)

\(\large m^2 - 3n^2\)

Answer 10

But how did you suddenly think of that? :O

Answer 11

\(xy = (m_1^2 -3n_1^2)(m_2^2 - 3n_2^2)\)

Answer 12

good question :)

from here :
http://math.stackexchange.com/questions/644861/if-both-integers-x-and-y-can-be-represented-as-a2-b2-4ab-prove-that

Answer 13

How did they think of it? :|
The thing is, I may be able to prove it like this but, i may lose points for directly getting this big a hint [directly] from the net. :P

Answer 14

since we're stuck on this for a while, and since we have exhausted all our options, getting hints from internet is okay

Answer 15

watch out for other cool methods if they exist but i dont think u can manage to put the product into original form, w/o change of coordinates

Answer 16

Yeah? I tried it for 2 hours straight...couldn't. :\
Okay, let me try it this method only.
I think we have to convert it to any possible binomial.

Answer 17

Wait a sec. How did they get that???

(x2−3y2)(u2−3v2)=(xu+3yv)2−3(xv+yu)2

Answer 18

expand out and see

Answer 19

\((x^2−3y^2)(u^2−3v^2) \)

Answer 20

I know it can be verified like that, but to get that, (xu+3yv)2−3(xv+yu)2 is impossible without knowing, you are going to get that. O.O

Answer 21

agree

Answer 22

@LastDayWork

Answer 23

This is the question with all the parts.

Answer 24

Sorry I couldn't reply last night...OS was down for me..

Everything else has been nicely done in stackexchange. As for the formula, it is obvious -
\[x^2- 3y^2=(x+y \sqrt3)(x-y \sqrt3)\]
For,
\[(x^2-3y^2)(u^2-3v^2)=(x+y \sqrt3)(x-y \sqrt3)(u+v \sqrt3)(u-v \sqrt3)\]
Club first and third term ; second and fourth term (in RHS).
Finally use the identity -
\[(a+b)(a-b)=a^2-b^2\]

Answer 25

np..

I actually got it from stackxchange only then.
I haven't slept all night though, so as it is morning here, I'll just go get some sleep. :P
Thanks btw. :D

Answer 26

I can barely calculate on my fingers if I stay awake all night..