Demonstrate that 4k + 3 cannot be perfect square

Question

I wonder if it can be mathematically demonstrated that

4k + 3 != n^2 where k,n are N (natural) numbers?

My daughter (6's grade) has a problem and the book's answer is just enumerating k elements

{3,7,11,19,23,27...}

saying that they cannot be perfect square.

Thank you,

score 1 · Answer 1 · answered Apr 21 '20 at 10:19

1

Hint: the square of an even number is $(2m)^2 = 4m^2 = 4k$.

The square of an odd number is $(2m+1)^2 = 4m^2 +4m+1 = 4k + 1$.

Can you finish?

answered Apr 21 '20 at 10:19

Deepak

Odd+even(here, $2$) is also odd. – jiten Nov 14 '21 at 07:08
@jiten I don't understand your comment. – Deepak Nov 14 '21 at 07:59
$4k+1$ is odd, $+2$ is odd too. Sorry, was wrong. You meant $(2m+1)^2+(\sqrt{2})^2 $ $\ne (2m+1)^2+2.(2m+1)\sqrt{2} + 2$. – jiten Nov 19 '21 at 00:55

1 Answers1