How to Prove by Mathematical Induction that 6 Divides (k^2+5)k for All k>=1?

[Xyius]

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!

 

[tiny-tim]

Hi Xyius!

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

show us what you've tried, and where you're stuck, and then we'll know how to help!

 

[Raphie]

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.

 

[alphachapmtl]

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

 

[cubzar]

(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.

 

[dextercioby]

You can prove it without mathematical induction, too

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

which is obviously divisible by 6.

 

[zetafunction]

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

f(x)=0mod(n)

so you try the ansatz x=imod(n) ehere 'i' runs over i=0,1,2,3,4,5,,...,n-1

 

[ramsey2879]

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

K(K^{2} + 5) = = K(K^{2} -1) + 6
= (K-1)*K*(K+1) + 6 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.

(K +1)^{3) +5(K+1) == K^3 + 3K^2 + 8K + 6