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Show 6 divides (n^2+5)n

by Xyius
Tags: divides
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Xyius
#1
Dec6-10, 05:24 PM
P: 437
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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tiny-tim
#2
Dec6-10, 05:35 PM
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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
#3
Dec6-10, 08:54 PM
P: 153
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
#4
Dec18-10, 10:16 AM
P: 81
Show 6 divides (n^2+5)n

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
#5
Jan14-11, 02:33 PM
P: 26
(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
#6
Jan15-11, 10:06 AM
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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.
zetafunction
#7
Jan16-11, 09:11 AM
P: 399
Quote Quote by alphachapmtl View Post
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
ramsey2879
#8
Jan17-11, 02:35 PM
P: 894
Quote Quote by Xyius View Post
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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