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Write 4420 as a sum of 2 squares? (No programs, use 2 square identity)

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I can't solve this for you, as my maths isn't good enough, but I can suggest a start: A number ending in 0 squared is a number ending in 00. A number ending in 1 or 9 squared is a number ending in 1. A number ending in 2 or 8 squared is a number ending in 4. A number ending in 3 or 7 squared is a number ending in 9. A number ending in 4 or 6 squared is a number ending in 6. A number ending in 5 squared is a number ending in 25. Thus you can eliminate some possibilities for the squares you search for.
@Cheese3.1416 Well, that's OK, a computer program might use this to narrow things down. But it is simpler to use the two square identity (a^2+b^2)(c^2+d^2) = (ac+bd)^2 + (ad-bc)^2
aka the Brahmagupta–Fibonacci identity

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Other answers:

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....
Maybe this is me being stupid, but all I can say about the two square identity is it gives me one equation with 4 unknowns.
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
\[2 \times 2 \times 5 \times 221\]
Yes, u could have bro0ken 221 into 13 *17
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....
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.
Damn. The moment I said this problem seemed recursive I should've realized I needed to do what you said 4 posts ago.
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...

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