Apteryx wrote:Thank you for that, it was very clearly expressed. And covered what I saw ( using my three brain cells till they squealed ) as the problem, you could never prove it correct, because of an infinity of potential numbers, right?

Not quite - it is possible to prove results for infinitely many numbers using mathematical induction. Suppose you want to prove some result for all natural numbers. Well, you could first show that the result is true for n = 1. Then you could assume the result is true for n = k and use this assumption to prove that the result is true for n = k + 1. Then the proposition being true for 1 implies it is true for 2, which implies it is true for 3, and so on.

This is the proof method that phlip's first paragraph outlines.

Here's an example:

Suppose that you wanted to prove that the number [imath](n^4 + 6n^3 + 11n^2 + 6n)[/imath] is always divisible by [imath]4[/imath], for all [imath]n[/imath].

Well, check that it's true for [imath]n = 1:[/imath]

[imath]1 + 6 + 11 + 6 = 24[/imath], which is divisible by [imath]4[/imath].

Now assume it's true for [imath]n = k[/imath]:

1) [imath]k^4 + 6k^3 + 11k^2 + 6k[/imath] is divisible by [imath]4[/imath].

What can we say about [imath]n = k + 1[/imath]? Well,

2) [imath](k + 1)^4 + 6(k + 1)^3 + 11(k + 1)^2 + 6(k + 1) = k^4 + 10k^3 + 35k^2 + 50k + 24[/imath]

If we subtract the first number from the second number, we get:

[imath](k^4 + 10k^3 + 35k^2 + 50k + 24) - (k^4 + 6k^3 + 11k^2 + 6k) = 4k^3 + 24k^2 + 44k + 24 = 4 * (k^3 + 6k^2 + 11k + 6)[/imath]

Notice that the difference of the two numbers is divisible by [imath]4[/imath], and (by our assumption) the first number was divisible by [imath]4[/imath], which means that the second number must also be divisible by [imath]4[/imath].

Hence we have proven with mathematical induction that [imath]n^4 + 6n^3 + 11n^2 + 6n[/imath] is divisible by 4 for all natural numbers.