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

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Cheese3.1416Best ResponseYou've already chosen the best response.1
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.
 one year ago

estudierBest ResponseYou've already chosen the best response.1
@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 + (adbc)^2
 one year ago

estudierBest ResponseYou've already chosen the best response.1
aka the Brahmagupta–Fibonacci identity
 one year ago

Cheese3.1416Best ResponseYou've already chosen the best response.1
You seem to know what you're doing better than I do. Is this a test?
 one year ago

estudierBest ResponseYou've already chosen the best response.1
Think of it as a tutorial...
 one year ago

Cheese3.1416Best ResponseYou've already chosen the best response.1
OK, challenge accepted. Whether it'll be completed is entirely another matter....
 one year ago

Cheese3.1416Best ResponseYou've already chosen the best response.1
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.
 one year ago

estudierBest ResponseYou've already chosen the best response.1
OK, a hint then: 5 is a factor of 4420 and you could write that as (2^2 +1^2)
 one year ago

Cheese3.1416Best ResponseYou've already chosen the best response.1
So now the problem becomes write 884 as the sum of two squares?
 one year ago

estudierBest ResponseYou've already chosen the best response.1
Well, yes, 5 is not the only factor....
 one year ago

Cheese3.1416Best ResponseYou've already chosen the best response.1
This is starting to look recursive...
 one year ago

estudierBest ResponseYou've already chosen the best response.1
First off, you better get a factorization of 4420
 one year ago

Cheese3.1416Best ResponseYou've already chosen the best response.1
\[2 \times 2 \times 5 \times 221\]
 one year ago

estudierBest ResponseYou've already chosen the best response.1
Yes, u could have bro0ken 221 into 13 *17
 one year ago

estudierBest ResponseYou've already chosen the best response.1
There is more than 1 answer so let's use 2^2 * 5 * 13 * 17 as the factorization
 one year ago

Cheese3.1416Best ResponseYou've already chosen the best response.1
I really need to make myself more intelligent.
 one year ago

estudierBest ResponseYou've already chosen the best response.1
That is just the usual factorization...
 one year ago

estudierBest ResponseYou've already chosen the best response.1
Small numbers , easy to put as sums of squares...
 one year ago

Cheese3.1416Best ResponseYou've already chosen the best response.1
I am now processing.
 one year ago

estudierBest ResponseYou've already chosen the best response.1
Save a little time (2^2+0^2) (2^2+1^2) (3^2+2^2) (4^2+1^2)
 one year ago

Cheese3.1416Best ResponseYou've already chosen the best response.1
There was no need to give me the answer.
 one year ago

estudierBest ResponseYou've already chosen the best response.1
That's not the answer....
 one year ago

Cheese3.1416Best ResponseYou've already chosen the best response.1
Well I can work it out myself.
 one year ago

Cheese3.1416Best ResponseYou've already chosen the best response.1
I'm about to check 130 and 34.
 one year ago

estudierBest ResponseYou've already chosen the best response.1
But u don't have to check anything, just use the 2 square identity....
 one year ago

Cheese3.1416Best ResponseYou've already chosen the best response.1
Ok, well if\[130=11^2+3^2\]\[34=5^2+3^2\]then I guess\[(33+15)^2+(3315)^2=4420\]
 one year ago

Cheese3.1416Best ResponseYou've already chosen the best response.1
Eh, stupid me. I should've realized instantly that's wrong.
 one year ago

estudierBest ResponseYou've already chosen the best response.1
Right idea, sort of...
 one year ago

estudierBest ResponseYou've already chosen the best response.1
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)
 one year ago

Cheese3.1416Best ResponseYou've already chosen the best response.1
I'm no good at being systematic.
 one year ago

estudierBest ResponseYou've already chosen the best response.1
I will give you a medal for effort (and negative brownie points for being unsystematic)
 one year ago

Cheese3.1416Best ResponseYou've already chosen the best response.1
That seems generous.
 one year ago

Cheese3.1416Best ResponseYou've already chosen the best response.1
Damn. The moment I said this problem seemed recursive I should've realized I needed to do what you said 4 posts ago.
 one year ago

estudierBest ResponseYou've already chosen the best response.1
Yes, procedure is simply a recursive algorithm when all is said and done...
 one year ago

Cheese3.1416Best ResponseYou've already chosen the best response.1
Thanks for the medal, though I personally feel a bit disappointed with myself.
 one year ago

estudierBest ResponseYou've already chosen the best response.1
S'OK, it's good practice...
 one year ago
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