Question

Prove that n^4 + 3n^2 + 2 is never square.
Where n is a normal number.

Answer 1

what is a "normal number"

Answer 2

natural, sorry :)

Answer 3

What do you mean square?

Answer 4

that it can never have the form of a^2, where a is a natural

Answer 5

are you stuck with the other?

Answer 6

yeah I don't know how to do that one

Answer 7

you mean it can never be a square number?

Answer 8

16 is a square , 25 is
17 , 18, 19, 20... not

Answer 9

Ok sorry :(

Answer 10

probably my bad :-)

Answer 11

factor it

Answer 12

(n^2 + 1) ( n^2 + 2)

Answer 13

ok you can do it by contradiction, assume that it does equal a square

Answer 14

factoring it does the job

Answer 15

how?

Answer 16

you just need 1 more step

Answer 17

think about inequalities

Answer 18

well n^2 + 1 is irreducible, and n^2 + 2 is irreducible

Answer 19

(n^2 + 1) ( n^2 + 2) is less than something and greater than another thing :-)

Answer 20

I know :)

Answer 21

cantor are you still thinking about it?

Answer 22

n^2 < n^2 + 1 < ...

Answer 23

close but not quite

Answer 24

think about what you need to make a square

Answer 25

why doesnt the fact that its irreducible finish it

Answer 26

Largo I updated the other one

Answer 27

I am not 100% sure of that, is it enough? maybe

Answer 28

but you can see that
(n^2 + 1)^2 < (n^2 + 1)(n^2 + 2) < (n^2 + 2)^2
so it lies between 2 consecutive squares, therefore it cannot be a square

Answer 29

yes that makes sense

Answer 30

oh ok, top and bottom. so a right triangle rotated about an axis produces a right cone