Show 6 divides (n^2+5)n

  • Thread starter Xyius
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In summary, Cubzar's proof says that if k = 1, then k^2+5 = 5 mod 6, and if k = 2, then k^2+5 = 3 mod 6, and so on.
  • #1
Xyius
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I need to prove by mathematical induction that 6 divides (k^2+5)k for all k>=1. It works for 1 but I am having trouble obtaining the k+1'st term from the k'th term. Can anyone help? Thanks!
 
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  • #2
Hi Xyius! :smile:

(try using the X2 icon just above the Reply box :wink:)

show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
  • #3
It might be helpful if you restate as k^3 + 5k. Immediately it will be apparent that 2 will divide all values. If 3 also divides all values, then you've got your divisibility by 6.
 
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  • #4
if k=0 mod 6, k^2+5 = 5 mod 6, (k^2+5)k = 0 mod 6
if k=1 mod 6, k^2+5 = 0 mod 6, (k^2+5)k = 0 mod 6
if k=2 mod 6, k^2+5 = 3 mod 6, (k^2+5)k = 0 mod 6
if k=3 mod 6, k^2+5 = 2 mod 6, (k^2+5)k = 0 mod 6
if k=4 mod 6, k^2+5 = 3 mod 6, (k^2+5)k = 0 mod 6
if k=5 mod 6, k^2+5 = 0 mod 6, (k^2+5)k = 0 mod 6
 
  • #5
(k^2+5)k=k^3+5k
((k+1)^2+5)(k+1)=(k^2+2k+6)(k+1)=k^3+3k^2+8k+6=(k^2+5)k + 3(k^2+k)+6.
3(k^2+k) is a multiple of 6 since k^2+k is even.
 
  • #6
You can prove it without mathematical induction, too

[tex] k\left(k^2 +5\right) =k\left[(k+1)(k+2)-3k+3\right]= k(k+1)(k+2)-3k(k-1) [/tex]

which is obviously divisible by 6.
 
  • #7
alphachapmtl said:
if k=0 mod 6, k^2+5 = 5 mod 6, (k^2+5)k = 0 mod 6
if k=1 mod 6, k^2+5 = 0 mod 6, (k^2+5)k = 0 mod 6
if k=2 mod 6, k^2+5 = 3 mod 6, (k^2+5)k = 0 mod 6
if k=3 mod 6, k^2+5 = 2 mod 6, (k^2+5)k = 0 mod 6
if k=4 mod 6, k^2+5 = 3 mod 6, (k^2+5)k = 0 mod 6
if k=5 mod 6, k^2+5 = 0 mod 6, (k^2+5)k = 0 mod 6

can this method be generalized for the POlynomial diophantine equation

[tex] f(x)=0mod(n) [/tex]

so you try the ansatz [tex] x=imod(n) [/tex] ehere 'i' runs over i=0,1,2,3,4,5,,...,n-1
 
  • #8
Xyius said:
I need to prove by mathematical induction that 6 divides (k^2+5)k for all k>=1. It works for 1 but I am having trouble obtaining the k+1'st term from the k'th term. Can anyone help? Thanks!
See Cubzar's posr. The other proofs don't seem to be using induction. You showed it for k = 1. I think that to prove by induction you, you would have to write k+1 where ever k appears in the equation and expand the equation, then show that the fact that the new equation is divisible by 6 follows directly from the fact that the origional equation is divisible by 6. Maybe you might get something like

[tex]K(K^{2} + 5) = = K(K^{2} -1) + 6[/tex]
= [tex](K-1)*K*(K+1) + 6[/tex] from which you could deduce that if it is true for k = n then it is obviously true for k = n + 1.

or otherwise show as Cubzar did that F(K+1)-F(K) [ i.e. 3K^2 + 3K + 6] is always divisible by 6 regardless of the value of K mod 6.

[tex](K +1)^{3) +5(K+1) == K^3 + 3K^2 + 8K + 6
 
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1. What does it mean for 6 to divide (n^2+5)n?

When we say that 6 divides (n^2+5)n, it means that 6 is a factor of (n^2+5)n, which means that (n^2+5)n is divisible by 6 without any remainder.

2. What is the significance of (n^2+5)n in this equation?

(n^2+5)n is the number that we are trying to determine if it is divisible by 6. It is a product of two terms, n^2+5 and n, and we are checking if this product is divisible by 6.

3. How do you prove that 6 divides (n^2+5)n?

To prove that 6 divides (n^2+5)n, we need to show that (n^2+5)n can be written as 6 multiplied by some other integer. This can be done by factoring (n^2+5)n and showing that one of the factors is 6.

4. Is (n^2+5)n always divisible by 6?

No, (n^2+5)n is not always divisible by 6. For example, if n=3, then (n^2+5)n will be equal to 42, which is divisible by 6. But if n=2, then (n^2+5)n will be equal to 18, which is not divisible by 6.

5. Can you provide an example of a value for n where 6 divides (n^2+5)n?

One example is n=4. When n=4, (n^2+5)n is equal to 108, which is divisible by 6 since it can be written as 6 multiplied by 18.

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