Question 847697: Prove or disporove. If (n^3)+2 is odd then n^2 is odd where n is an element of all integers.
Found 2 solutions by jim_thompson5910, swincher4391:
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! This isn't a formal proof, but it's just the basic outline (to get you thinking in the right direction)
If n^3 + 2 is odd, then n^3 must be odd. Why? Because the +2 makes you jump from even to even or from odd to odd. An example would be 7+2 = 9. Notice how we go from 7 to 9 when we increase by 2.
So we know n^3 is odd. It turns out that if you cube any even number, you get an even number. Similarly, if you cube any odd number, you get an odd number. So because n^3 is odd, this must mean that n has to be odd.
We know n is odd. So n = 2k + 1 for some integer k. Square this to get n^2 = 4k^2 + 4k + 1 ---> n^2 = 2(2k^2 + 2k) + 1 ----> n^2 = 2m + 1 for some integer m (m = 2k^2 + 2k). This shows us that if n is odd, then n^2 is odd.
Let me know if this helps or not.
Answer by swincher4391(1107)  (Show Source):
You can put this solution on YOUR website! Assume (n^3) + 2 is odd.
Subtracting 2 does not change the parity of our integer.
So n^3 is odd.
If n^3 is odd, then let's look at the cube root of odd perfect cubes.
cube root of 1 = 1 <--- odd (implies that n^3 that ends in 1, will have an n that ends in 1)
cube root of 343 = 7 <--- odd (implies that n^3 that ends in 3 will have an n that ends in 7)
cube root of 125 = 5 <---- odd (implies that n^3 that ends in 5 will have an n that ends in 5)
cube root of 27 = 3 <---- odd (implies that n^3 that ends in 7 will have an n that ends in 3)
cube root of 729 = 9 <---- odd (implies that n^3 that ends in 9 will have an n that ends in 9)
Notice that no matter what odd n^3, we'll get an odd n.
What's left to show is that if n is odd, then n^2 is odd.
1^2 = 1 <-- odd (if n ends in 1, n^2 will end in 1)
3^2 = 9 <-- odd (if n ends in 3, n^2 will end in 9)
5^2 = 25 <--- odd (if n ends in 5, n^2 will end in 5)
7^2 = 49 <--- odd (if n ends in 7, n^2 will end in 9)
9^2 = 81 <--- odd (if n ends in 9, n^2 will end in 1)
Given an odd n, we know that we will have an odd n^2.
This completes the proof.
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