Equation Manipulation

Date: 4/13/96 at 20:3:29
From: Anny
Subject: SAT problem

Hi.  I'm taking the Princeton Review Course right now and my 
instructor and I are having a problem with a certain question.  
We know the answer, but just can't figure out how it works.  So, 
here goes...

Number 10 on Section 6 of a practice SAT:

If m and n are integers, which of the following could be equal to 
{mr + ns}^2 for all values of r and s?

(A) 2r^2 + 4rs + 2s^2
(B) r^2 + 4rs + s^2
(C) r^2 + 2rs + 4s^2
(D) r^2 + 2rs - s^2
(E) r^2 - 2rs + s^2

If you could help out it would be greatly appreciated.  I hope you 
guys have better luck than we did!

Date: 4/14/96 at 15:9:31
From: Doctor Syd
Subject: Re: SAT problem

Hello!

Let's first write out (mr + ns)^2:

(mr + ns)^2 = m^2r^2 + 2mnrs + n^2s^2

Right?

Okay, So, this means the coefficient of r^2 must be a square of an 
integer, right?  Well, this rules out part (a) since 2 isn't the 
square of an integer.  For b through e, suppose each of the given 
answers could be equal to (mr + ns)^2, and figure out what the 
values of m and n would  then have to be.  You'll get a 
contradiction for all but one of them.

For instance, let's do (b):  

If r^2 + 4rs + s^2 = m^2r^2 + 2mnrs + n^2s^2 

then we must have that m^2 = 1 and n^2 =1, and 2mn = 4.

But this can't be since m^2 = 1 and n^2 = 1 implies that m is plus 
or minus 1 and n is plus or minus 1, so mn is plus or minus 1 
(considering all of the cases), so 2mn is plus or minus 2 which is 
not equal to 4.  

See if you can try out c-e and figure out which one doesn't give 
you a contradiction!  Hope this helps.  Write back if you have 
more questions.

-Doctor Syd,  The Math Forum