Question

If A is the set of all numbers a^2+4ab+b^2, where a and b are both integers, prove that A is closed under multiplication.

I've tried a few different things, but the algebra always gets out of hand. Any help would be appreciated!

Answer 1

so we want to show the set of numbers of the form p is closed under multiplication where p= a^2 +4ab + b^2 and a,b are integers. Now to show closure we need two elements. call them p1 and p2 . show that p1 * p2 is of the form m^2 + 4m*n + n^2 for some integers m,n

Answer 2

Got there, and multiplied it out, but I couldn't see a way forward.

Answer 3

Also, wouldn't we need to fix m and n instead? The choice of m and n is arbitrary, but we should need to fix them to prove closure.

Answer 4

now you need two integers for each number. so let p1 = r^2 + 4rs + s^2
and p2 = u^2 + 4u*v+ v^2

Answer 5

*Sorry, didn't see you were still typing

Answer 6

then p1 * p2 = (r^2 + 4rs + s^2)*(u^2 + 4u*v + v^2)

Answer 7

=
r^2*u^2+4*r^2*u*v+r^2*v^2+4*r*s*u^2+16*r*s*u*v+4*r*s*v^2+s^2*u^2+4*s^2*u*v+s^2*v^2

Answer 8

Looks like what I got

Answer 9

can we simplify this

Answer 10

we can rewrite this sum
r^2*u^2+4*r^2*u*v+r^2*v^2+4*r*s*u^2+16*r*s*u*v+4*r*s*v^2+s^2*u^2+4*s^2*u*v+s^2*v^2
=
r^2u^2 + r^2v^2 + s^2u^2 + s^2v^2 + 4( r^2uv + rsu^2 + rsv^2 + s^2uv + 4rsuv)

Answer 11

That's what I got also (using different variables, so it's taking me a couple minutes to check)

Answer 12

WolframAlpha says that bit inside the parentheses isn't factorable, so that's where I got stuck.

Answer 13

Unless some of those terms belong with the square terms (the a^2 and b^2 terms)

Answer 14

yes this looks pretty ugly ,
you can say that the product of two integers is an integer, and keep using this result

Answer 15

but i dont see the exact way to get it in the form
m^2 + 4mn + n^2

Answer 16

Me either. I generated the first few numbers, and it seems to check out.

Answer 17

there is an identity you can use
a^2+b^2+4ab=(a+2b)^2-3b^2

Answer 18

that might be easier to deal with

Answer 19

I tried these guys:
\[(a+b)^2 +2ab\]\[2(a+b)-(a^2+b^2)\]

Answer 20

so using the identity, we have
p1 * p2
= (r^2 + 4rs + s^2)*(u^2 + 4u*v + v^2)
= [ (r + 2s)^2 - 3s^2] [ (u + 2v)^2 - 3v^2]

Answer 21

Looks good

Answer 22

so using the identity, we have
p1 * p2
= (r^2 + 4rs + s^2)*(u^2 + 4u*v + v^2)
= [ (r + 2s)^2 - 3s^2] [ (u + 2v)^2 - 3v^2]
= (r+2s)^2*(u+2v)^2 - 3s^2*(u+2v)^2 - 3v^2*(r+2s)^2 + 9s^2v^2

Answer 23

ok i found a new identity, this one will work :)

Answer 24

if f(x,y) = x^2 + 4xy + y^2
Then
f(a,b)* f(c,d)=f(ac-bd,ad+4bd+bc)

Answer 25

do you see how it works now?

Answer 26

f(r,s) * f(u,v) = f( ru - sv , rv + 4sv + su )
so those will be our m and n

Answer 27

(r^2 + 4rs + s^2)*(u^2 + 4u*v + v^2)
= (ru-sv)^2 + 4 (ru - sv) *(rv +4sv + su) + (rv + 4sv + su)^2

Answer 28

In your initial setup for the function, how did you get your x to be ac-bd?

Answer 29

now let m = (ru-sv) , n = (rv + 4sv + su)

Answer 30

sorry, i didnt understand

Answer 31

f(a,b)* f(c,d)=f(ac-bd,ad+4bd+bc)
Where did you get the ac-bd part?

Answer 32

but it works

Answer 33

Identity for which part? I see that it's supposed to be x, but
? = ac-bd

Answer 34

break it down into parts
remember the rule is f(x,y) = x^2 + 4xy + y^2
so...
f(a,b) = a^2+4ab+b^2
f(c,d) = c^2+4cd+d^2
f(ac-bd,ad+4bd+bc) = (ac-bd)^2 + 4(ac-bd)(ad+4bd+bc) + (ad+4bd+bc)^2

Answer 35

Ah, alright, back with you now.

Answer 36

so we are basically using the identity
(a^2 +4ab + b^2)(c^2+4cd+d^2)= (ac-bd)^2 + 4(ac-bd)(ad+4bd+bc) + (ad+4bd+bc)^2

Answer 37

the lefthand side can be rewritten using the shorthand f(a,b) * f(c,d)
the right side can be written using the shorthand f( ac-bd, ad+4bd + bc)

Answer 38

I see the identity now :) . Awesome. How did you find that?

Answer 39

if the f(a,b) stuff confuses you , you can ignore it. the main thing is the identity
(a^2 +4ab + b^2)(c^2+4cd+d^2)= (ac-bd)^2 + 4(ac-bd)(ad+4bd+bc) + (ad+4bd+bc)^2
we can write the product of two numbers of the set A as again a member of the set of A.

Answer 40

No, it doesn't. Just thought that you derived it somehow and I didn't see where it comes from. I can see that it works, but I'm wondering how it was derived in the first place.

Answer 41

oh sorry about that :)

Answer 42

Nah, it's awesome. I have been making 0 headway with this problem. How did you find that identity?

Answer 43

yeah i dont see an easy way to derive this
I found it here
http://math.stackexchange.com/questions/644861/if-both-integers-x-and-y-can-be-represented-as-a2-b2-4ab-prove-that

see the comment by heropup

Answer 44

Thanks :)

Answer 45

scroll down to the second answer

Answer 46

you can use a more abstract approach, using norms
what class are you in

Answer 47

I'm a middle school teacher (I was a math major), but a friend was asking me about a few problems for some Stanford math camp. They're allowed to use whatever they want as long as they say where they got help.

Thanks a lot for the help!

Answer 48

ahh, ok :)

Answer 49

I tutor math for a living, this was a fun problem :)

Answer 50

oh it looks like there is another way to do this, or perhaps it is the same

Answer 51

I'm pretty sure that ac-bd thing came from the idea of a normed vector space. I've seen it before, but I'm not sure where. I think it's used a lot in programming...

Answer 52

we have two identities:
a^2+b^2+4ab=(a+2b)^2-3b^2
And :
(x^2-3y^2)(u^2-3v^2)=(xu+3yv)^2-3(xv+yu)^2

Answer 53

so to do this requires two substitutions, keeping track of everything is kind of tricky though

Answer 54

Yeah, I'm going to throw this at her and make her do it :P

Answer 55

r^2 + 4rs + s^2 = ( r + 2s ) ^2 - 3s^2
u^2 + 4uv + v^2 = (u + 2v)^2 - 3v^2

Answer 56

:D

Answer 57

i was doing this before, but i didnt finish it

Answer 58

I'll let her play with it and I may repost if she doesn't get anywhere.

Answer 59

These problems are pretty fun. There's one where I'm wondering if they gave enough information.

Answer 60

sure :)

Answer 61

but of course to get this identity simply from thin air probably isnt going to happen, so you have to use those other more well known tricks/identities

Answer 62

Yeah, that looks like what they did on here with the Brahmagupta-Fibonacci Identity

Answer 63

are you from USA?

Answer 64

yes, VA
You?

Answer 65

yes from new york

Answer 66

Nice. Get a lot of snow these past few weeks?

Answer 67

yes , hehe