Lcm of k,k+1,k+2 where k≡3(mod 4) ?

[kingwinner]

Find the least common multiple of k,k+1, and k+2 where k≡3(mod 4).

Some ideas:
k≡3(mod 4) => k=3+4z for some z E Z
So k is odd, k+1 is even, and k+2 is odd.
Also, this question asks about the lcm of THREE numbers, so I don't think the the Euclidean algorithm (which only gives the gcd or lcm of TWO numbers) will help us in this question.

So how can we solve this?
Thanks for any help!

 

[JSuarez]

What is the gcd of k, k+1 and k+2? If you know it (and you can), the lcm is immediate.

 

[kingwinner]

Right now, I really don't have any way to calculate the gcd of THREE numbers. My textbook talks about the Euclidean algorithm, but all examples and problems are dealing with the gcd of TWO numbers.

Another thing is we don't know what those numbers k, k+1, k+2 actually are...

 

[Hurkyl]

Yes you do -- they are k, k+1, and k+2.

If you only know how to compute pairwise LCM's, then seek to reduce the problem to that.

 

[JSuarez]

What can you say about the common divisors of consecutive integers? Notice that (k+1)-k = 1 and (k+2)-k = 2 and k is odd. By the way, gcd(a,b,c) = gcd(gcd(a,b),c).

 

[kingwinner]

What can you say about the common divisors of consecutive integers? Notice that (k+1)-k = 1 and (k+2)-k = 2 and k is odd. By the way, gcd(a,b,c) = gcd(gcd(a,b),c).

How can we prove that gcd(a,b,c) = gcd(gcd(a,b),c)?

I know that gcd of two consecutive integers = gcd(k,k+1)=1, i.e. two consecutive integers are always relatively prime, but I have no idea about the gcd or lcm of three consecutive integers...

 

[JSuarez]

How can we prove that gcd(a,b,c) = gcd(gcd(a,b),c)?

Well, suppose that d is a common divisor of a,b and c, then it will also be a common divisor of...

I know that gcd of two consecutive integers = gcd(k,k+1)=1

How do you prove this?

 

[ramsey2879]

Well, suppose that d is a common divisor of a,b and c, then it will also be a common divisor of...



How do you prove this?

Yep if 3 divides each of a,b,c it will also divide each of a and b! A common divisor of a and b must equal or divide the gcd(a,b).

 

[ramsey2879]

What is the gcd of k, k+1 and k+2? If you know it (and you can), the lcm is immediate.

You can find the lcm of a,b,c by multiplying a,b,c together and dividing that result by gcd(a,b,c) twice and by gcd(a,b)/gcd(abc) and gcd(a,c)/gcd(a,b,c) and gcd(b,c)/gcd(a,b,c) each once. You know how to multiply, add and subtract algebraic numbers don't you.

 

[kingwinner]

k is odd, k+1 is even, and k+2 is odd

I know that two consecutive integers are always relatively prime, so gcd(k,k+1)=1 and gcd(k+1,k+2)=1.

With k and k+2 odd, is it always true that gcd(k,k+2)=1? If so, how can we prove it?

If the above are true, then k, k+1, k+2 are pairwise coprime, then it seems intuitively believable to me that lcm(k,k+1,k+2)=k(k+1)(k+2). But I can't figure out a way to prove this rigorously...

Can someone help me, please?
Thank you!

 

[JSuarez]

With k and k+2 odd, is it always true that gcd(k,k+2)=1? If so, how can we prove it?

Yes. What does the equality (k+2)-k = 2 tell you about the possible values of gcd(k,k+2)?

k, k+1, k+2 are pairwise coprime, then it seems intuitively believable to me that lcm(k,k+1,k+2)=k(k+1)(k+2)

Think of the lcm in terms of the prime factorizations of k,k+1,k+2.

 

[kingwinner]

OK, now I have proved that k, k+1, k+2 are pairwise coprime.

But I still can't figure out how to prove that lcm(k,k+1,k+2)=k(k+1)(k+2).

Can somebody provide some help, please?

 

[tt2348]

so you have GCD(K,K+1,K+2)=1, just use the definition of LCM, which is
LCM(a1,a2,a3...)=(a1*a2*a3*a4...)/(GCD(a1,a2,a3...)

 

[kingwinner]

so you have GCD(K,K+1,K+2)=1, just use the definition of LCM, which is
LCM(a1,a2,a3...)=(a1*a2*a3*a4...)/(GCD(a1,a2,a3...)

But that formula is not true...
$$\text{lcm}(2,4,8)\cdot\gcd(2,4,8) = 2\cdot 8\neq 2\cdot 4\cdot 8$$

So how can we PROVE that k, k+1, k+2 are pairwise coprime => lcm(k,k+1,k+2)=k(k+1)(k+2)??

Thanks.

 

[JSuarez]

Let $k = p_{1}^{e_1}p_{2}^{e_2}\cdots p_{n}^{e_n}$, $k+1 = q_{1}^{f_1}q_{2}^{f_2}\cdots q_{n}^{f_n}$ and $k + 2 = r_{1}^{g_1}r_{2}^{g_2}\cdots r_{n}^{g_n}$ be the prime factorizations of k, k+1 and k + 2. They don't have any common prime factors; so, what are the factors that must appear on the lcm?

 

[kingwinner]

Let $k = p_{1}^{e_1}p_{2}^{e_2}\cdots p_{n}^{e_n}$, $k+1 = q_{1}^{f_1}q_{2}^{f_2}\cdots q_{n}^{f_n}$ and $k + 2 = r_{1}^{g_1}r_{2}^{g_2}\cdots r_{n}^{g_n}$ be the prime factorizations of k, k+1 and k + 2. They don't have any common prime factors; so, what are the factors that must appear on the lcm?

So I guess we just multiply all of them.

But k=1 and negative k's don't have prime factorizations, how can we prove for these cases?
Also, LCM(a1,...an) itself is defined to be positive, but the a1,...,an can be integers. So will this change our answer? Maybe the lcm should be |k(k+1)(k+2)|?

Thanks!

 

[JSuarez]

So I guess we just multiply all of them.

Exactly.

But k=1 and negative k's don't have prime factorizations, how can we prove for these cases?

Remember that, if k=1, then any positive integer is a multiple of k, so this doesn't affect the result. Second, negative integers have prime factorizations: you just have to put a minus sign in front.

Maybe the lcm should be |k(k+1)(k+2)|?

It is: the expression for two numbers is:

$$\rm{lcm}\left(a,b\right)=\frac{\left|ab\right|}{\rm{gcd}\left(a,b\right)}$$