SOLUTION: How to prove {{{n^2-n+14}}} is divisible by 2 using mathematical induction given it's true for all natural numbers (n)?

   
We are to prove:

Pn:  n²-n+14 is divisible by 2 

induction proof:
 
First let's see what Pk+1 would be:
 
[That's always the best first thing to do.  Before you start an induction
proof,  you should calculate Pk+1 to see where you're headed]:

To do that, replace n by k+1 in n²-n+14 to see what the Pk+1 is,

for that is what we are going for, and if we have that beforehand,

we'll know when we have arrived and the proof is finished.

Substituting k+1 for n in n²-n+14, we have

(k+1)²-(k+1)+14 = k²+2k+1-k-1+14 = k²+k+14.

So now we have Pk+1, which is where we'll be headed

Pk+1:  k²+k+14 is divisible by 2 

Now that we know what Pk+1 will have to be, we know where we're going, 
and we'll know we have arrived if and when we get that. 

 
So now we can start the proof:

P1:  1²-1+14 is divisible by 2

That's true because 1²-1+14 = 1-1+14 = 14, which is divisible by 2 
because (2)(7) = 14  

Assume k is some integer for which k is true.  We know there is at least
one such value, because we just proved P1.

Pk:  k²-k+14 is divisible by 2 
 
We look at the expression in Pk+1, that we're going for, and realize that 
the difference between the expression in Pk+1 and the expression in Pk is
(k²+k+14)-(k²-k+14) = k²+k+14-k²+k-14 = 2k 

2k is a multiple of 2, and we know that if we add two multiples of 2 we
get another multiple of 2.

So assuming 

Pk:  k²-k+14 is divisible by 2

     k²-k+14 + 2k is also divisible by 2
So

     k²+k+14 is divisible by 2

and that is Pk+1

So the proof is finished.

Now we show that it is finished:

So since P1 is true, 1 is a possible value for k, and therefore
P1+1 or P2 is true,  so P1 proves P2, P2 proves P3, P3 proves
P4, etc., etc., ad infinitum.

No matter how large an integer we have proved that P is true, 
we have proved that the next integer will also be one in which P 
is also true.

Edwin