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aka the Brahmaguptaâ€“Fibonacci identity

You seem to know what you're doing better than I do. Is this a test?

Think of it as a tutorial...

OK, challenge accepted.
Whether it'll be completed is entirely another matter....

:-)

OK, a hint then:
5 is a factor of 4420 and you could write that as (2^2 +1^2)

So now the problem becomes write 884 as the sum of two squares?

Well, yes, 5 is not the only factor....

This is starting to look recursive...

First off, you better get a factorization of 4420

\[2 \times 2 \times 5 \times 221\]

\[221=14^2+5^2\]

Yes, u could have bro0ken 221 into 13 *17

\[20=2^2+4^2\]

There is more than 1 answer so let's use
2^2 * 5 * 13 * 17 as the factorization

I really need to make myself more intelligent.

That is just the usual factorization...

Small numbers , easy to put as sums of squares...

I am now processing.

Save a little time
(2^2+0^2) (2^2+1^2) (3^2+2^2) (4^2+1^2)

There was no need to give me the answer.

You spoiler.

That's not the answer....

Oh.

Well I can work it out myself.

I'm about to check 130 and 34.

But u don't have to check anything, just use the 2 square identity....

Ok, well if\[130=11^2+3^2\]\[34=5^2+3^2\]then I guess\[(33+15)^2+(33-15)^2=4420\]

Try again...

Eh, stupid me. I should've realized instantly that's wrong.

Right idea, sort of...

More systematic wouold be to apply the identity 3 times to
(2^2+0^2) (2^2+1^2) (3^2+2^2) (4^2+1^2)

I'm no good at being systematic.

I will give you a medal for effort (and negative brownie points for being unsystematic)

That seems generous.

Yes, procedure is simply a recursive algorithm when all is said and done...

Thanks for the medal, though I personally feel a bit disappointed with myself.

S'OK, it's good practice...