Question

Use the method of proof of exhaustion to prove that if "a" is not a multiple of 5, then (a^4 -1) is a multiple of 5.

Answer 1

Proof by exhaustion? Isn't that just proof by cases?

Answer 2

That it is. Exhaust all possibilities. Hit it, tiger :P

Answer 3

In that case, you can use modulo arithmetic.

I can explain it to you, but it'd be a shame not letting the expert, Terence, explain :-)

Answer 4

No point for flattery, Parth. This is good experience... :P

Answer 5

But, who can really resist the lure of a good non-calculus question. @sedighn
I stand ready...
At your signal... we begin ^_^

Answer 6

Hahaha ready when you are :)

Answer 7

That's the spirit :)
Suppose a is not divisible by 5. Then it must leave a remainder when divided by 5, am I right?

Answer 8

Yess

Answer 9

What remainders could it possibly leave? Remember, we're exhausting all possible cases.

Answer 10

oops sorry my internet died. you could have remainder 1,2,3 or 4, am i right?

Answer 11

Stumped?

Answer 12

Ok. That's good.

Answer 13

So, there are four possible cases, but before that, I want you to note this property
\[\large (5h-k)^4=(5h)^4-4\cdot(5a)^{3}k+6\cdot(5h)^{2}k^2-20hk^3+k^4\]

Answer 14

You can solve for this, but for now, take my word for it. Catch me so far? Don't worry, this is just a once-over.

Answer 15

I seee, ok

Answer 16

So, I want you to notice, that if we divide this part by 5\[\large (5h-k)^4=\color{red}{(5h)^4-4\cdot(5a)^{3}k+6\cdot(5h)^{2}k^2-20hk^3}+k^4\]

Answer 17

It's not going to leave a remainder, right? That part is quite divisible by 5.

Answer 18

ahhhh yes

Answer 19

So the only thing that MIGHT leave a remainder is the part \(\large k^4\)

Answer 20

rightt

Answer 21

So, suppose a leaves a remainder of 1, when divided by 5. That means
\[\large a=5h+1\]right?

Answer 22

for some integer h.

Answer 23

ohh I see

Answer 24

and then you substitute that into a^4 -1 and factor out the 5 to show that it is a multiple of 5, right?

Answer 25

and then repeat for a=5h+2, a=5h+3 and a=5h+4 ?

Answer 26

Yes. You'll find, as we have shown, that the remainder of
\[\large a^4=(5h+1)^4\]leaves a remainder of \(\large 1^4=1\) when divided by 5.

Answer 27

So, \[\large a^4 = 5n+1\]

Answer 28

yessss I think I've got it, thank you so much!!