A)

It factorises to X, X-1 and X+1, one of which must be divisible by three

C

Nice one Elfffi.  Want a harder one?

F) Some bricks

Elffi is awesome!

You can knock out c-e by having x=1, and b by having x=2.

Scep Tick also a nice practical way to solve.  Elfffi's answer was very concise. 

None of them

A) fails where x=1 as 0 is not divisible by 3

Good spot - should have probably said natural number rather than whole.  

If x² + 3x + 1 = 0, what is the value of x² + (1/x)²?

0 is divisible by 3.  And indeed by every number.

yeah ok i will accept that, depending on definition of divisibility, there is a good argument for 0 being divisible by all numbers with the answer always being 0

personally i do not find that v useful practically, but it’s undeniably a strong argument so i will allow berb’s challenge as it is

0⁰ = 1.  Just to confuse things.

heh

i’m gonna go ahead and put that in the debatable category too

Scep Tick's approach doesn't show that A) is divisible by 3 for all x.

The answer to the OP could be none.

To be fair, I didn't put in my post that b) to e) are not valid.

Answer to the second question is 7

I used a long, brute force approach. There is probably a more elegant solution

No, I don’t have a clue why ROF insists on presenting it upside down. Perhaps it’s something to do with the way Apple stores images in the iPad photo library.

True enough.  Pal Erdos used to offer less money for counter-examples than he did for proofs...

Solid work Elfffi, you take the crown as ROF's maths nerd.

The neat solution is:

• x² - 3x +1 = 0
• therefore x -3 + 1/x = 0
• therefore x + 1/x = 3
• squaring both sides, (x + 1/x)² = 9
• opening brackets x² + 2 + 1/x² = 9
• therefore x² + 1/x² = 7

Elf dude have u tried writing the right way up?  Makes this shiz loads easier