Question

show that the product of sum of squares of two integers can be represented as sum of squares of two integers

Answer 1

\((x^2+y^2)(m^2+n^2) = u^2 + w^2\)

Answer 2

Are you allowed to use complex numbers?

Answer 3

yes anything is okay...

Answer 4

I recall seeing two different ways to show this, one with and one without complex numbers to show how much easier they can make some things... I don't know if I quite recall, but it's something like either taking the real part of something or maybe like this: \[\LARGE a a^*bb^*=cc^*\] Let me think a second to see if I can figure it out.

Answer 5

Oh nice, i think you're using |z1||z2| = |z1z2|

Answer 6

Oh I know: \[\LARGE (aa^*)(bb^*)=(ab)(ab)^*\]

Does that make sense @ganeshie8

Answer 7

yes i just made it up, so its possible...

Answer 8

no I just solved it, see?

Answer 9

it does make sense to me and it looks real cute :)
i think we need to agree that magnitude of prodcuts is same as product of magnitudes

Answer 10

\[\LARGE a=x+iy \\ \LARGE b=m+i n \] \[\LARGE (aa^*)(bb^*)=(x^2+y^2)(m^2+n^2)\]\[\LARGE (ab)(ab)^*=(xm-yn)^2+(my+xn)^2\]

Answer 11

ok like this :-
(x^2+y^2)(n^2+m^2)=(x+y)(x-iy)(n+m)(n-im)=(xm+xn+ym+yn)(xm-inx-iym+yn)
= number * its conjocate = something ^2 +something ^2

Answer 12

I guess we'll just have to get a divorce, sorry.

Answer 13

BABY KAI DOESNT HOLD BACK! WOOOT

Answer 14

wait not conjocate xD
bhaha
what i mean is like this
(xm+xn+ym+yn)(xm-inx-iym+yn)=((xm+yn) + (xn+ym) )) ((xm+yn) -i (xn+ym) ))= (xm+yn)^2 + (xn+ym)^2

Answer 15

same idea as kai :D

Answer 16

Actually there's sort of an interesting recursive nature to this, if you can express 1 sum of two squares as two sums of two squares multiplied together, then that means you can express it as 3 sums of two squares multiplied together.\[\LARGE (a^2+b^2)(c^2+d^2)(e^2+f^2)=(g^2+h^2)\]

But we can go further, that means two squares added together can be represented by an infinite amount of two squares added together right??? \[\LARGE (a^2+b^2)... \infty ...(w^2+x^2)=(y^2+z^2)\]

Answer 17

In complex numbers this playing around amounts to this I guess idk can we do calculus on this somehow lol\[\LARGE (aa^*)(bb^*)...(zz^*)=(ab...z)(ab...z)^*\]

Answer 18

|z1| |z2| |z3| ... = |z1z2z3...|
so yeah it works xD

Answer 19

well instead of making it sum of squares , make it n power xD
and it might work also (just conjecture )

Answer 20

but but
(4+9)(100+49)=

Answer 21

|(2+3i)(10+7i)|

Answer 22

how did u create this ?

Answer 23

it reminds with something but i cant remember what it is xd

Answer 24

oh yes , i remembered any even number u can write it as a sum of 2 primes

Answer 25

\[(x^2+y^2)(m^2+n^2)=(xm-yn)^2+(my+xn)^2\]

Answer 26

thats the formula in the end of kai's work above

Answer 27

Here's an idea.\[\LARGE \ln(x^2+y^2)=\sum_{n=0}^\infty \ln(a_n^2+b_n^2)\]

Answer 28

kai the Hexagon :P