1. Arithmetic Prove

IMPORTANT: Number and explain (in english) your steps.

Prove that, every square number has either the form 4q or 4q + 1 for some integer q.

Use formats of sqrt(x) for √x and x^k for xk

Added to this post by the OP after the reply by Mr F:

I have this

We are trying to prove the statement, x^2 = 4n or 4n + 1 for some integer n. If x is even, then x = 2k for some integer k, so x^2 = 4k^2. Setting n = k^2 gives us an integer n such that x^2 = 4n, and we are done. If x is odd, then x = 2k + 1 for some integer k. Thus, x^2 = (2k+1)^2 = 4k^2 + 4k + 1. Letting n = 4(k^2 + k), we have x^2 = (2k+1)^2 = 4n + 1. Since k is an integer, n = 4(k^2 + k) is also an integer

2. Originally Posted by smh745
IMPORTANT: Number and explain (in english) your steps.

Prove that, every square number has either the form 4q or 4q + 1 for some integer q.

Use formats of sqrt(x) for √x and x^k for xk
Step1: What have you tried?

Spoiler:
Steps 2 - : Note that square numbers can be generated by taking the product of two consecutive even or odd numbers and adding 1.