Prove: 2 is a factor of (n^4-n+4)

Question

I need full steps written out using mathematical induction please help!

 

Teacher lists three steps as

 

prove true for n=1

 

assume true for n=k

 

Prove true for n=k+1

Answer 1

Prove that n⁴-n+4 is even when n=1:   1⁴-1+4 = 4

 

Assume that n⁴-n+4 is even when n = k  (Assume that k⁴-k+4 is even.  So, k⁴-k+4=2r, for some integer, r.)

 

   Show that n⁴-n+4 is even when n=k+1.  (Show that (k+1)⁴-(k+1)+4 is even)

 

      (k+1)⁴-(k+1)+4 = k⁴+4k³+6k²+4k+1-k-1+4 =(4k³+6k²+4k) +(k⁴-k+4)

 

                                                                        = 2(2k³+3k²+2k) + 2r (by assumption)

 

                                                                        = 2(2k³+3k²+2k+r)

 

                                                                         Since r and k are integers, 2k³+3k²+2k+r is an integer.

 

     So, (k+1)⁴+(k+1)+4 is even.

 

By induction, n⁴+n+4 is even for all integers n ≥ 1. 

 

Answer 2

 

n=1 works out to  1-1+4 = 4  which has a factor of 2.

 

Assume n⁴-n+4 is even for n=k  (and that expression in terms of k is simply k⁴-k+4)

 

Now for n=k+1 we get   (k+1)⁴-(k+1)+4   which needs to be shown to have a factor of 2.

 

(k+1)⁴ -(k+1) + 4 =  (k²+2k+1)²-(k+1) + 4

= (k⁴ + 4k³ + 6k² + 4k + 1) - (k+1) + 4

= k⁴ + 4k³ + 6k² + 3k + 4

 

Now look at the difference between k⁴ -k+4  and  this last expression:

(k⁴+4k³+6k²+3k+4) - (k⁴-k+4)

=  4k³+6k²+4k

= 2(2k³+3k²+2k) 

 

This last expression has a factor of 2 (i.e. it is even).

 

So we know k⁴-k+4 has a factor of 2 (by induction hypothesis),  and if we add the even number 2(2k³+3k²+2k) to it, we get the value for n=k+1.  Therefore n⁴-n+4 always has a factor of 2. 

 

 

 

Answer 3

In order for 2 to be a factor of a number, that number must be even and a multiple of 2.

 

n=1  makes the expression even and a multiple of 2.              1 - 1 + 4 = 4

 

If n=1 makes the expression even and a multiple of 2 and n=k, then k also makes the expression even and a multiple of 2.

 

 

For   n = k + 1:

 

(k + 1)⁴ - (k + 1) + 4 = (k + 1)⁴ - k + 3

                                 = (1 + 1)⁴ - 1 + 3

                                 = 16 + 2

                                 = 18

 

18 is even and a multiple of 2.