# Correct me if I am wrong

1. Sep 17, 2005

### Jarvis Bull Dawg

Prove that of 25 is subtrated from the square of an odd integer greater than 5, the resulting number is always divisible by 8.

Solution:

Let S (n) = n2 – 25
For n = 1, S (1) = (1 )2 – 25 = -24 which is clearly divisible by 8.
Thus the first condition is satisfied as S(1) is true.

Let us assume that S(n) is true for n = 2k+1 Belonging to Odd integers greater than 5, that is,

S(2k+1) = (2k+1)2 - 25 is divisble by 8
= 4k2 + 4k + 1 -25
= 4k2 + 4k - 24
= 4(k2 + k - 6)
= 4(k-2)(k+3) ------ (A)

We can see that (A) is clearly divisible by 8, condition being that k is an odd integer which is bigger than 5. Since both conditions are satisfied, hence by mathematical induction we have proven that S(n) is divisble by 8 for all integers bigger than 5.

2. Sep 17, 2005

### ComputerGeek

umm, you are not exactly correct. you are setting n = to 2k + 1. while that is denoting an odd integer, that is what n should be and you have not shows for n+1 yet which should be 2k + 3.

however, induction doe snot require that you use that form. you can simply say n for all n = 2k+1, now showing for n+1 and just use n+1 to show it rather than 2k+3

3. Sep 17, 2005

### shmoe

Why is your base case n=1 when you're trying to prove this for odd integers greater than 5?

You don't use your induction hypothesis here, and you don't need to. Induction is not necessary.

n>5, not k.

4. Sep 17, 2005

### Jarvis Bull Dawg

True. Thank you

What should I use?

Next thing, so should I use k+1 rather than 2k+1? Advice?

5. Sep 17, 2005

### shmoe

2k+1 is good, n is an odd integer after all. You don't know if k is even or odd though.

use what you've got:

n^2-25
=(2k+1)^2 - 25
= 4k^2 + 4k + 1 -25
= 4k^2 + 4k - 24
= 4(k^2 + k - 6)
= 4(k-2)(k+3) ------ (A)

Show A is divisible by 8. 2 cases if you like, k even or k odd.

Actually the n>5 bit isn't necessary, it's just to keep n^2-25 positive, the concept of divisibility still makes sense for negative integers.

By the way you can use Latex to make exponents $$n^{2}$$ <-click to see how, or please use a ^ to denote exponentiation as I've done above, as "n2" looks like another variable or the variable n times 2.

6. Sep 18, 2005

### rachmaninoff

Odd integers greater than 5: 7, 9, 11...
(5+2), (5+4), (5+6)...

general form: (5+2n), n=1,2,3...

So, what number are you looking at? (5+2n)^2 minus something? Do some algebra here...

Until you're done. (hint: factor out '2's when you can...)

There's no need for induction. If you must, start with n=7, not n=1 (factoring negative integers is uncool).

Last edited by a moderator: Sep 18, 2005
7. Sep 18, 2005

### Mithal

I think it is easier solved this way

(n^2)-25 divides 8 is equivalent to

(n^2)-1 divides 8 because 24 divides 8

now we have

(n-1) (n+1) divides 8

n= 2 m + 1 ( n is odd as suggested in the problem)

(2 m ) (2 m + 2 ) = 4 m (m+1)

if m is even we are done , if m is odd then (m+1) is even .Both cases prove

divisibility by 8

8. Sep 18, 2005

### Jarvis Bull Dawg

Here is another one, but seriously I have no idea what this is.

Observe that the last two digits of $$7^2$$ are 49, the last two digits of $$7^3$$ are 43, the last two digits of $$7^4$$ are 01, and the last two digits of $$7^5$$ are 07. Prove that the last two digits of $$7^201$$ are 07.

9. Sep 18, 2005

### shmoe

what are the last two digits of $$7^6$$? $$7^7$$? Notice a pattern?

Suggestion:$$7^{4}=100k+1$$ for some integer k. Use this to show that the last two digits of $$7^{4n}$$ are 01 for any natural number n.

10. Sep 18, 2005

### Jarvis Bull Dawg

Yes I do notice the recurring pattern, but I didn't get your suggestion.

11. Sep 18, 2005

### shmoe

Knowing $$7^{4}=100k+1$$ you should be able to prove you can write $$7^{4n}=100m+1$$ where m is some integer (for example use the binomial theorem).

This will prove that the last two digits of $$7^{200}$$ are 01. (Since you're only interested in this case you could really take n=50 above, but it doesn't really simplify things)

By the way, you're not familiar with modular arithmetic are you? While not necessary to do problems like this, it does simplify things.

12. Sep 18, 2005

### Jarvis Bull Dawg

Edited for errors

Last edited: Sep 19, 2005
13. Sep 18, 2005

### Jarvis Bull Dawg

Hope yo get it. I solved it for the 201 term. By B-Theorem

14. Sep 19, 2005

### Mithal

I tried to do it this way
By binomial theorem I find the last two terms are
(10-3) ^ 201 = (201) (10) (3 ^ 200 )- (3) ^ 201
or
(3 ^ 200) (2010 - 3)
or
(9^100) (2000+7) =((10-1)^100 )( 7) since we are concerned with the last two digits
or
7(1000 t +1)
7000 t +7
This proves the last two digits are 07

15. Sep 20, 2005

### Werg22

You're solutions are perhaps correct but you fail to proove the generality of this statement (for any therm). The true solution is much simpler...

Any odd number greater than 5 can be writtent under the form 5+x, where x is even. Following the statement:

(5 + x)^2 - 25

25 + x^2 + 2(5)x - 25

x^2 + 2(5)x.

Since x is even, let x=2y

(2y)^2 + 2(2)(5)y

4y^2 + 4(5y)

4(y^2 + 5y)

y^2 + 5y is even for any integrer. So let y^2 + 5y=2a

4(2a)

8a

This is the proof that the solution for (5 + x)^2 - 25, where x is even is always a factor of 8.

Last edited: Sep 20, 2005