Prove: The square of a number that is one more than a multiple of 3 is also one
more than a multiple of 3.
n +
(3n + 1)2 = [ ? ]n? +
1([ ]n2 +
=
)n) +
- 1 more than a multiple of 3

Let 3n + 1 denote the "number" in question. The claim is that

(3n + 1)² = 3m + 1

for some integer m.

Now,

(3n + 1)² = (3n)² + 2 (3n) + 1²

… = 9n² + 6n + 1

… = 3n (3n + 2) + 1

… = 3m + 1

where we take m = n (3n + 2).

rotation point l

step-by-step explanation: