
Why is the last digit of n^5 equal to the last digit of n? [video] - iamzlatan
https://www.youtube.com/watch?v=ZQUTV9or98s
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air7
This reminds me why I don't like induction proofs. Although sound, almost by
definition no insight can be obtained from it for either why the proposition
is true, or why the proof itself is valid.

Here's another proof I just though of that doesn't use induction: Let's prove
that x^5-x is divisible by 10.

    
    
      x^5-x = x(x^4-1) = x(x^2+1)(X^2-1)
    

We want to prove that among these 3 multiplied elements, we can find the
factors 2 & 5\. 2: If x is even, it's x. If x is odd, it's the other ones. 5:
Consider the following table of : x | x^2 (mod 5)

    
    
      x | x^2 | Which element is divisible by 5
      -------------------------------------------
      0    0       x
      1    1       x^2-1
      2    4       x^2+1
      3    4       x^2+1
      4    1       x^2-1
    

so we get that in all cases there is a factor of 5 as well.

Not sure this provides more insight into why the proposition is true, but IMO
this proof _feels_ more valid.

~~~
eeereerews
You can do it without a table by factoring

x (x^2-1) (x^2+1) =

x (x-1) (x+1) (x^2 - 4 + 5) =

x (x-1) (x+1) ((x-2)(x+2) + 5)

Now one of the five consecutive numbers (x-2), (x-1), x, (x+1), (x+2) must be
divisible by 5, so the whole is as well.

I agree with you about induction.

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jwilk
TL;DW: proof by induction.

Basis is trivial: 0⁵ = 0.

Inductive step:

Assume that n⁵ - n is divisible by 10.

We'll show that (n + 1)⁵ - (n + 1) is divisible by 10, too.

(n + 1)⁵ - (n + 1)

= n⁵ + 5n⁴ + 10n³ + 10n² + 5n + 1 - n - 1

= (n⁵ - n) + 10(n³ + n²) + 5n(n³ + 1)

Notice that:

* (n⁵ - n) is divisible by 10, as per the inductive hypothesis;

* 10(n³ + n²) is obviously divisible by 10;

* either n or n³+1 is even, so 5n(n³ + 1) is divisible by 10.

QED.

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SamReidHughes
Dupe:
[https://news.ycombinator.com/item?id=15572593](https://news.ycombinator.com/item?id=15572593)

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jwilk
Alternative proof:

From Fermat's little theorem: n^5 ≡ n (mod 5).

Therefore (n^5 - n) is divisibe by 5.

(n^5 - n) is always even, so it's also divisble by 10.

