Math Is Fun Forum

miyazaki tezuka

Member

Registered: 2006-11-08

Posts: 2

NUMBER THEORY! pls. help me answer this. =)

uhhmm.. this is our assignment in number theory and i am having a hard time answering this. can someone please help me?
well, it is more on proving.

1. Show that if 2a+1 is a multiple of 5, then 14a^2 + 19a - 19 is a multiple of 25.
2. If a and b are odd integers, then 8 divides (a^2 - b^2).
3. Prove that if a divides b and a + b = c, then a divides c.
4. Prove that if a divides c and a + b = c , then a divides b.
5. If n is an odd integer, then 24 divides n (n^2 - 1).

to those who will help me answer this,, thank you thank you thank you!

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: NUMBER THEORY! pls. help me answer this. =)

For 3 & 4, first think about your times tables from 3rd grade.
If 5 divides 35, then 5 + 35 is also in the 5's tables.
And for #4, if 4 divides 24, then 4 should divide 20, which is 4 less, since they are still in the 4's tables.

---

igloo myrtilles fourmis

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: NUMBER THEORY! pls. help me answer this. =)

1. Preform long division by dividing 14a^2 + 19a - 19 by 2a + 1.  You will get a remainder, make sure you leave this as an integer (don't divide in polynomial fractions).  So you will have P(a)*(2a + 1) + Remainder = 14a^2 + 19a - 19.  What needs to be seen is every piece of the left is divisible by 25.

2.  Really simple.  Start with two odd numbers: 2k+1 and 2l + 1, square and subtract.  Then factor out the 4, and note that an even plus and even is even, and an odd plus an odd is even.  So is the stuff inside even or odd?  What does this mean you can do?

3. We know that a | a, and a | b.  So does a | (a + b)?

4. Same question, just subtract a from both sides.

5. Multiply it all out and factor it down to 4k(2k+1)(k+1).  Note that 24 = 2^3 * 3.  We already got 2^2.  Show that there is another 2 factor in there by first assuming k is even, then assuming k is odd.  Now all you need to show is that 3 either divides 2k + 1 or k+1, no matter what your choice of k is.

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."