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

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

Setting $\displaystyle n=6k+5$ we have immediately…

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

Regards

Hello, Cabouli!

Here are a few of them . . .

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

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

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

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

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

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

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

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

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

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


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

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: .$\displaystyle a^3\,\text{ and}\,(a+1)^3$

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

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

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

Thank you for all your help.

Cheers
Cabouli