Number theory integer proofs

**Cabouli** · March 10th 2009, 03:23 AM

Some number theory problems:
1. Show that every integer of the form 6k+5 must also be of the form 3k+2
2. Show that the sqaure of any odd integer is of the form 8k+1
3. Prove that the cube of any integer can be written as one of 9k-1,9k or 9k+1
4. Prove that the sum of the squares of two odd integers cannot be a perfect square.
5. Prove that the difference of two consectutive cubes is odd.

Any help would be gratefully accepted.

Cheers
Cabouli

**chisigma** · March 10th 2009, 06:24 AM

Show that every integer of the form 6k+5 must also be of the form 3k+2…

Setting $n=6k+5$ we have immediately…

$n=5 \mod 6 = 2 \mod 3$

Regards

**Soroban** · March 10th 2009, 06:31 AM

Hello, Cabouli!

Here are a few of them . . .

2. Show that the square of any odd integer is of the form $8k+1$

An odd integer has the form: . $N = 2a+1\,\text{ for some integer }a.$

Its square is: . $N^2 \:=\:(2a+1)^2 \:=\:4a^2 + 4a + 1$

We have: . $N^2 \:=\:4\underbrace{a(a+1)} + 1$

Note that $a$ and $a+1$ are consecutive integers.
So one of them is even and the other is odd.
Hence: . $a(a+1)$ is an even integer, $2k,$

And we have: . $N^2 \;=\;4(2k)+1 \;=\;8k+1$

4. Prove that the sum of the squares of two odd integers cannot be a perfect square.

If an integer is even, $2a$ ,
. . its square is: . $4a^2$ , a multiple of 4.

If an integer is odd, $2a+1$ ,
. . its square is: . $4a^2 + 4a + 1 \:=\:4a(a+1) + 1$ , one more than a multiple of 4.

Hence, the square of an integer has the form $4k\,\text{ or }\,4k+1$

The sum of the squares of two odd integers is:
. . $(2a+1)^2 + (2b+1)^2 \:=\:4a^2 + 4a + 1 + 4b^2 + 4b+1 \:=\: 4(a^2+b^2+a+b) + 2$

This is two more than a multiple of 4 . . . It cannot be a square.

5. Prove that the difference of two consectutive cubes is odd.

The two consecutive cubes are: . $a^3\,\text{ and}\,(a+1)^3$

Their difference is: . $(a+1)^3-a^3 \:=\:a^3 + 3a^2 + 3a + 1 - a^3 \:=\:3a^2 + 3a + 1$

$\text{We have: }\;3\underbrace{a(a+1)} + 1$
The product of two consecutive integers is even, $2k$

$\text{So we have: }\;3(2k) + 1 \:=\:\underbrace{6k}_{\text{even}} + 1 \:=\: \text{odd}$

**Cabouli** · March 10th 2009, 10:45 PM

Thank you for all your help.

Cheers
Cabouli

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Number theory integer proofs

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