Expert: Paul Klarreich - 10/6/2008

Question
Ok Paul, you have been amazing so far! Here are two more for you to have fun with. (1) If (A) is an even integer and (B) is an odd integer, then 4 does not divide(a^2+2b^2). (2) For every integer m such that 2 divides m and 4 does not divide m, there exists no integer x and y for which x^2+3y^2=m.

For

Answer
Questioner:   Pete
Category:  Calculus
Private:  No
 
Subject:  Advanced Mathematics-Proofs
Question:  Ok Paul, you have been amazing so far! Here are two more for you to have fun with.
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Hi, Pete,

(1) If (A) is an even integer and (B) is an odd integer, then 4 does not divide (a^2 + 2b^2).

Write  a = 2n, and  b = 2m + 1   << standard way to write even and odd.  In fact, this is how you prove even and odd:

If you can write  a = 2n you proved it is even.
If you can write  b = 2m + 1, you proved it is even.
..........

a^2 + 2b^2 = 4n^2 + 2(4m^2 + 4m + 1)

= 2(2n^2 + 4m^2 + 4m + 1)

If that is divisible by 4, then

2(2n^2 + 4m^2 + 4m + 1) = 4p, and

2n^2 + 4m^2 + 4m + 1 = 2p

2(n^2 + 2m^2 + 2m) + 1 = 2p

So the LHS is of the form  2q + 1, therefore ODD,
while the RHS is of the form  2p, therefore EVEN.
..................................
Next case:

(2) For every integer m such that 2 divides m and 4 does not divide m, there exists no integer x and y for which x^2+3y^2=m.

[In other words, x^2 + 3y^2 is EITHER divisible by 4 or is ODD.]

Suppose  x,y both odd:

x = 2j + 1,  y = 2k + 1
x^2 = 4j^2 + 4j + 1
y^2 = 4k^2 + 4k + 1

x^2 + 3y^2 = 4j^2 + 4j + 1 + 12k^2 + 12k + 3 = 4(stuff) ==> Divisible by 4.

Suppose  x,y both even:  Then x^2, y^2 are both divisible by 4, and so is any combination.

Suppose  x is even,  y is odd:

x = 2j,  y = 2k + 1

x^2 = 4j^2
y^2 = 4k^2 + 4k + 1

x^2 + 3y^2 = 4j^2 + 12k^2 + 12k + 3 =
4j^2 + 12k^2 + 12k + 2 + 1 =
2(stuff) + 1 ==> ODD.

Finally suppose  x odd, y even:

x = 2j + 1,  y = 2k

x^2 = 4j^2 + 4j + 1
y^2 = 4k^2

x^2 + 3y^2 = 4j^2 + 4j + 1 + 12k^2 = 4j^2 + 4j + 12k^2 + 1 = 2(stuff) + 1, ODD.

That's it.