Question

a) Prove that the square of any odd multiple of 3 is the difference of two rectangular no.
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Answer 1

Answer:

"Prove the square of an odd multiple of 3 is the difference of two triangular numbers."

So I know an odd multiple of 3 is always in the form of

(3*(2c + 1)), so that squared is 36c2 + 36c + 9

I also know that a triangular number is in the form of n(n+1)/2. Looking online I found the same problem but with an extra sentence saying "in particular, show that for any natural number n, so that [3(2n + 1)]2ˆ = t9n+4 - t3n+1."

I can do it by cheating and working backwards from there

t9n+4 - t3n+1 =ˆ [(9n + 4)(9n + 5)]/2 - [(3n + 1)(3n + 2)]/2

= (72n2 + 72n + 18)/2 ˆ = 36c2 + 36c + 9

But I can't for the life of me show that 36c2 + 36c + 9 is [(9n + 4)(9n + 5)]/2 - [(3n + 1)(3n + 2)]/2. I'm stuck at this point (72n2 + 72n + 18)/2. How could I possibly (even know to) decompose

(72n2 + 72n + 18)/2

into

[(9n + 4)(9n + 5)]/2 - [(3n + 1)(3n + 2)]/2?