# Show 6 divides (n^2+5)n

by Xyius
Tags: divides
 P: 383 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!!
 PF Patron HW Helper Sci Advisor Thanks P: 25,480 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!
 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.
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
 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.
 HW Helper Sci Advisor P: 11,717 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.
P: 399
 Quote by 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
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
P: 891
 Quote by 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!!
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.

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

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