Question

show that \(m^2+n+2\) and \(n^2 +4m\) are not perfect squares simultaneously

Answer 1

limited to integers? whole numbers?

Answer 2

sry yes limited to whole numbers

Answer 3

*positive integers

Answer 4

oooh! hey I can actually do this one!

Answer 5

i bet you can :)

Answer 6

Number theory proofs are not my strong suit, so I get excited when I can as it is not a usual occurrence.

Answer 7

So, I would approach this by first defining what it would mean if each respectively was a perfect square

Answer 8

here is a hint given book

if \(k^2\) is a perfect square (it is!) then the next perfect square occurs after \(2k+1\) integers

that means for \(m^2+n+2\) to be a perfect square, we must have \(n+2 \ge 2m+1\)

Answer 9

I think you can pull a wlog m>n

Answer 10

true

Answer 11

as for the hint, do you see where they got that from?

Answer 12

not you dan haha

Answer 13

not quite.. im still unsure of this prblem actually

Answer 14

ok so let's look at it from the beginning. Do you agree that if \(m^2+n+2=k^2 \) is a perfect square then there exists some k>2 for it?

Answer 15

we can actually make that sharper I think

Answer 16

a^2 - b^2 = (a+b)(a-b)

a^2 = m^2+n+2
-b^2 = -(n^2+4m)

m^2 +n +2 -n^2 -4m

(m^2-4m) - (n^2 -n) +2

(m^2-4m+4) - (n^2 -n+1/4) +2-4+1/4

(m-2)^2 - (n-1/2)^2 -7/4 = (m-2+n-1/2)(m-2-n+1/2) -7/4

not sure where im going with this thought .... just thinking tho

Answer 17

ugh, but we need to prove the perfect squares layout first in a lemma