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So i want to prove that every perfect square is of the form 4k or 4k+1, can someone tell me how to do this. Really need help with this

Syed Naqi
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2 Answers2

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Any natural number is either even or odd.
So any natural number can be represented as $2n$ or $2n+1$ where $n \in \mathbb{N}$.

Now, $(2n)^2=4n^2$ which is of the form $4k$ where $k \in \mathbb{N}$
AND
$(2n+1)^2=4n^2+4n+1=4n(n+1)+1$ which is of the form $4k+1$ where $k \in \mathbb{N}$.

Hope this helps.

SchrodingersCat
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  • Thanks for that quick response guys really helpful and Thankyou – Syed Naqi Nov 13 '15 at 17:16
  • @SyedNaqi You're welcome. You can accept my answer by clicking the "tick" mark beside my answer. – SchrodingersCat Nov 13 '15 at 17:17
  • it won't let me after 7 min i can accept an answer. also btw how did you do this because I don't know how to do these discrete math problem every time I try to I always get stuck with these logical question – Syed Naqi Nov 13 '15 at 17:19
  • Actually, for any such discrete math problem, you need to start your problem from the basics. Consider what kind of numbers are suitable for your problem, the representation I used here is useful in many problems. And the more you practise such problems, the more you will learn. – SchrodingersCat Nov 13 '15 at 17:21
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For every square $a^2$ there are two possible cases:

  • Case 1: $a^2$ is even. Then, it is the square of an even number $a = 2n \Rightarrow a^2 = 4n^2 = 4k$

  • Case 2: $a^2$ is odd. Then, it is the square of an odd number $a = 2n + 1 \Rightarrow a^2 = 4n^2 +4n + 1= 4(n^2 + n) + 1 = 4k + 1$

nabla
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