find the squares of 3n, 3n+1, 3n+2?
Answers
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Step-by-step explanation:
First we look at n=1 and try to do a proof by induction.
Obviously 1^2 = 1 = 3*0 + 1 works.
That means for some x we have x^2=3N or x^2=3N + 1.
Now we look at (x+1)^2 = x^2 + 2x + 1
Using the case x^2 = 3N we can just ignore that term and we’re left with 2x+1. Now x mod 3 can only be 0 because x^2 is divisible by 3, that means x must be divisible by 3, and therefore x mod 3 is 0. Then (x+1)^2 = x^2 + 2x + 1, we know x^2 = 3N for some N. Then we’re left with 3N + 2x + 1, but x can be written as 3k for some k (because x is divisible by 3). We’re then left with (x+1)^2 = 3N + 3k + 1 = 3(N+k) + 1.
Now if x^2 = 3N + 1 then x mod 3 can either be 1 or 2. If it’s 1 then (x+1)^2 = x^2 + 2x + 1 = 3N + 1 + 2x + 1, and x can be written as 3k+1 for some k. We’re left with (x+1)^2 = 3N + 2(3k+1) + 2 = 3N + 6k + 3 + 1 = 3(N + 2k + 1) + 1.
If x mod 3 is 2 then repeat the same process and we’re left with
(x+1)^2=3(N + 2k +2)
Because adding one got either a term of form 3N or 3N+1 we conclude that each perfect square must be of that form.
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