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

[CharlesLin]

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 calcualte 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.

 

[kaliprasad]

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 calcualte 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

 

[CharlesLin]

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!

 

[kaliprasad]

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.

 

[Greg]

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.

 

[I like Serena]

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)

 

[Greg]

Yes! I was thinking "minimum even factor of X".

- - - Updated - - -

Which would, of course, be 2.

 

[CharlesLin]

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?

 

[MarkFL]

If you use the hint provided by kaliprasad in his first post, you then obtain:

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

What do you get when you factor that?

 

[CharlesLin]

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?

 

[I like Serena]

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.

 

[MarkFL]

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:

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

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?

 

[CharlesLin]

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?

 

[MarkFL]

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 $8$.

 

[I like Serena]

...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 $8$.

 

[MarkFL]

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 $\gcd(m,n)=\gcd(6,4)=2$. :)

 

[CharlesLin]

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

 

[MarkFL]

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.

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:

\(\displaystyle 6=2\cdot3\)

\(\displaystyle 4=2^2\)

and so we see the largest factor common to both is 2.

 

[CharlesLin]

ok I understand,

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

 

[MarkFL]

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:

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

and

\(\displaystyle \gcd(6,4)=2\)