prove that the square of some integer is of the form 4n, 4n+1 for some integer n
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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∈N.
Now, (2n)^2=4n^2 which is of the form 4k where k∈N
AND
(2n+1)^2=4n^2+4n+1=4n(n+1)+1 which is of the form 4k+1 where k∈N.
Hope this helps.
@skb
So any natural number can be represented as 2n or 2n+1 where n∈N.
Now, (2n)^2=4n^2 which is of the form 4k where k∈N
AND
(2n+1)^2=4n^2+4n+1=4n(n+1)+1 which is of the form 4k+1 where k∈N.
Hope this helps.
@skb
RabbitPanda:
Thnx 4 brainlist
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