MHB Finding the Greatest EVEN Factor of X: Solving the GCD of m,n=2

  • Thread starter Thread starter CharlesLin
  • Start date Start date
  • Tags Tags
    even Gcd
Click For Summary
SUMMARY

The discussion centers on determining the greatest even factor of the expression X = 6m² + 4n², given that the greatest common divisor (GCD) of m and n is 2. Participants explored various pairs (m, n) that satisfy this GCD condition, ultimately concluding that the greatest even factor of X is 8, derived from the factorization of X as 8(3p² + 2q²). The values of p and q do not need to be consecutive, and the focus should be on finding pairs that yield an odd result for 3p² + 2q² to ensure that 8 is a consistent factor.

PREREQUISITES
  • Understanding of GCD (Greatest Common Divisor) and its implications.
  • Familiarity with polynomial expressions and factorization techniques.
  • Knowledge of even and odd integers in mathematical contexts.
  • Basic algebraic manipulation skills, particularly with quadratic expressions.
NEXT STEPS
  • Study the properties of GCD and how they apply to polynomial expressions.
  • Learn about factorization techniques for quadratic expressions, specifically in the form of ax² + bx + c.
  • Explore the implications of odd and even integers in mathematical proofs and problem-solving.
  • Investigate the relationship between GCD and factorization in algebraic expressions.
USEFUL FOR

Mathematicians, students studying algebra, and anyone interested in number theory or polynomial factorization techniques will benefit from this discussion.

CharlesLin
Messages
16
Reaction score
0
this one got me thinking for a while it starts like this:

X=6m2+4n2

and Greatest Common Divisor(GCD) of (m,n)=2

what is the greatest EVEN number that must be a factor of X

I started this question by thinking what they asked, the gratest number that is a factor of X then I need to calculate X
I know that de GCD of m,n is 2, then I just have to find these two numbers, but the list of two numbers with 2 as a GCD start growing

(2,2)
(2,2)
(2,6)
(2,8)
(2,10)...

Therefore, I would like to know if you have a better way of finding "m" and "n", or how I can be sure that the a pair of those that I found is the correct.
 
Mathematics news on Phys.org
CharlesLin said:
this one got me thinking for a while it starts like this:

X=6m2+4n2

and Greatest Common Divisor(GCD) of (m,n)=2

what is the greatest EVEN number that must be a factor of X

I started this question by thinking what they asked, the gratest number that is a factor of X then I need to calculate X
I know that de GCD of m,n is 2, then I just have to find these two numbers, but the list of two numbers with 2 as a GCD start growing

(2,2)
(2,2)
(2,6)
(2,8)
(2,10)...

Therefore, I would like to know if you have a better way of finding "m" and "n", or how I can be sure that the a pair of those that I found is the correct.

gcd of m and n is 2. so we have m =2p and n = 2q with p and q are co-primes so they do not share a common factor. can you proceed from here
 
Ok following that hint that you gave me,

m=2p, n=2q

I try giving values to p and q. I toke in consideration that p and q are co-primes, in other words, they must consecutive.

p=2, q=3

x=6(16)+4(36)
x= 96+144
x=1050

then I have that 1050 is X but how can I know that the values that I chose for p and q are the appropiate?
If i continue form here knowing that the value of X=1050, I know that the gratest EVEN number that is a factor of 1050 ia 1050 Am I right? then 1050 is my answer!
 
CharlesLin said:
Ok following that hint that you gave me,

m=2p, n=2q

I try giving values to p and q. I toke in consideration that p and q are co-primes, in other words, they must consecutive.

p=2, q=3

x=6(16)+4(36)
x= 96+144
x=1050

then I have that 1050 is X but how can I know that the values that I chose for p and q are the appropiate?
If i continue form here knowing that the value of X=1050, I know that the gratest EVEN number that is a factor of 1050 ia 1050 Am I right? then 1050 is my answer!

you do not need x. you need to know the highest power of 2 that divides x.
further computation of x is incorrect. it is 240.
 
p and q need not be consecutive. That being said, try keeping the factor of 6m2 + 4n2 which is a power of 2 at a minimum. So you have m = 2p and n = 2q with p and q coprime odd numbers.
 
Last edited:
Erm... from the expression of x we can tell that x is even.
Isn't the greatest even factor of x then x itself? (Wondering)
 
Yes! I was thinking "minimum even factor of X".

- - - Updated - - -

Which would, of course, be 2. :o
 
ok, I feel more confused...
what I'm looking is X to be able to answer the question of What's the greatest EVEN number that is a factor of X.

the hint you gave me is that

I shoul find p and q which are co-primes but they aren't consecutive...

my question would be p=1 and q=3 could be a possibility?if so I would have X=168, however someone mentioned that I don't need to find X. I really don't understand why? and how do you know that p and q are co-prime and odd values?
 
If you use the hint provided by kaliprasad in his first post, you then obtain:

$$X=6\left(2p\right)^2+4\left(2q\right)^2=24p^2+16q^2$$

What do you get when you factor that?
 
  • #10
well once you factor that expression...

$$8\left(3{x}^{2}+2{y}^{2}\right)$$

buth then, how do I know what is the answer?
 
  • #11
CharlesLin said:
ok, I feel more confused...
what I'm looking is X to be able to answer the question of What's the greatest EVEN number that is a factor of X.

Can you clarify the problem statement?

The greatest even factor (that is not necessarily prime) of X is X itself.
The greatest even prime factor is $2$.
We currently seem to be heading for the greatest power of $2$ that is a factor of X, but that is not what the problem statement is asking.
 
  • #12
CharlesLin said:
well once you factor that expression...

$$8\left(3{x}^{2}+2{y}^{2}\right)$$

buth then, how do I know what is the answer?

Yes, we get:

$$X=6\left(2p\right)^2+4\left(2q\right)^2=24p^2+16q^2=8\left(3p^2+2q^2\right)$$

Now, if we can find at least one ordered pair $(p,q)$ such that the factor $3p^2+2q^2$ is odd, then what may we conclude?
 
  • #13
so we got to this point...
given this equation

$x=6{m}^{2}+4{n}^{2}$

what is the greatest even number that MUST be a Factor of X?

taking m=2p and n=2q

we have

$x=6{\left(2p\right)}^{2}+4{\left(2q\right)}^{2}$

x=$x=6{\left(2{p}^{2}\right)}+4{\left(4{q}^{2}\right)}$

$x=24{p}^{2}+16{q}^{2}$

$x=8\left(3{p}^{2}+2{q}^{2}\right)$

now I have to look for two numbers p and q that give me an odd number for $3{p}^{2}+2{q}^{2}$ is that right?

I choose p=1, q=3

but I still don't see the answer using (1,3) x=42 is this the answer?
 
  • #14
What I was getting at is that if we can find a $(p,q)$ such that $3p^2+2q^2$ is odd (which you did, and in fact as long as $p$ is odd then $3p^2+2q^2$ will be odd) then we may conclude that the largest even number that will always be a factor of $X$ is $8$.
 
  • #15
MarkFL said:
...the largest even number that will always be a factor of $X$ is $8$.
I think that should be: the largest power of 2 that is a factor of any $X$ is $8$.
 
  • #16
I like Serena said:
I think that should be: the largest power of 2 that is a factor of any $X$ is $8$.

The problem asked for the largest even factor of $X$...which does happen to be a power of 2 given that $\gcd(m,n)=\gcd(6,4)=2$. :)
 
  • #17
thank you very much guys I think I almost got it.

so to find the greatest even number that must be a factor of X I don't need to fin X. Right?

then I don't need to find p and q. But then MarkFL how do you know that (m,n) is (6,4)=2
 
  • #18
CharlesLin said:
so to find the greatest even number that must be a factor of X I don't need to fin X. Right?

$X$ is given as a function of $m$ and $n$ and so it isn't one specific value, but varies according to the definition given.

CharlesLin said:
then I don't need to find p and q. But then MarkFL how do you know that (m,n) is (6,4)=2

We are told that $\gcd(m,n)=2$ and we can compute $\gcd(6,4)=2$ since the prime factorizations of these numbers are:

$$6=2\cdot3$$

$$4=2^2$$

and so we see the largest factor common to both is 2.
 
  • #19
ok I understand,

but how do you know that (n,m) is 6,4 and no other number?
 
  • #20
CharlesLin said:
ok I understand,

but how do you know that (n,m) is 6,4 and no other number?

Sorry, I didn't mean to imply that $m$ and $n$ had any set values, I only meant:

$$\gcd(m,n)=2$$

and

$$\gcd(6,4)=2$$
 

Similar threads

MHB Is $\gcd(n, n+2) = 1$ or $2$ for any integer $n$?
  • · Replies 5 ·
Replies
5
Views
2K
Undergrad Questions regarding Kurepa's Conjecture
  • · Replies 1 ·
Replies
1
Views
2K
Undergrad Understanding the Results of gcd(x,n)
  • · Replies 8 ·
Replies
8
Views
2K
MHB Can GCD of m and n be forced to equal 1 for given equations and conditions?
  • · Replies 3 ·
Replies
3
Views
3K
MHB Prove that CnXCm isomorphic to Cgcd(m,n)XClcm(m,n)
  • · Replies 3 ·
Replies
3
Views
3K
Finding GCD with Fibonacci: Base Case
  • · Replies 1 ·
Replies
1
Views
2K
2x2 matrix with factor group elements
  • · Replies 2 ·
Replies
2
Views
2K
Undergrad Micromass' big October challenge
  • · Replies 125 ·
5
Replies
125
Views
20K
Undergrad Is this proof of the least common multiple theorem in number theory valid?
  • · Replies 1 ·
Replies
1
Views
1K
MHB Solve √(x^2+3x+7)-√(x^2-3x+9)+2=0
  • · Replies 5 ·
Replies
5
Views
2K