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Write 4420 as a sum of 2 squares?
(No programs, use 2 square identity)
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Cheese3.1416
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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.
estudier
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@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
estudier
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aka the Brahmagupta–Fibonacci identity
Cheese3.1416
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You seem to know what you're doing better than I do. Is this a test?
estudier
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Think of it as a tutorial...
Cheese3.1416
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OK, challenge accepted.
Whether it'll be completed is entirely another matter....
estudier
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:-)
Cheese3.1416
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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.
shubhamsrg
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*
estudier
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OK, a hint then:
5 is a factor of 4420 and you could write that as (2^2 +1^2)
Cheese3.1416
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So now the problem becomes write 884 as the sum of two squares?
estudier
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Well, yes, 5 is not the only factor....
Cheese3.1416
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This is starting to look recursive...
estudier
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First off, you better get a factorization of 4420
Cheese3.1416
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\[2 \times 2 \times 5 \times 221\]
Cheese3.1416
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\[221=14^2+5^2\]
estudier
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Yes, u could have bro0ken 221 into 13 *17
Cheese3.1416
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\[20=2^2+4^2\]
estudier
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There is more than 1 answer so let's use
2^2 * 5 * 13 * 17 as the factorization
Cheese3.1416
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I really need to make myself more intelligent.
estudier
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That is just the usual factorization...
estudier
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Small numbers , easy to put as sums of squares...
Cheese3.1416
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I am now processing.
estudier
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Save a little time
(2^2+0^2) (2^2+1^2) (3^2+2^2) (4^2+1^2)
Cheese3.1416
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There was no need to give me the answer.
Cheese3.1416
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You spoiler.
estudier
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That's not the answer....
Cheese3.1416
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Oh.
Cheese3.1416
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Well I can work it out myself.
Cheese3.1416
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I'm about to check 130 and 34.
estudier
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But u don't have to check anything, just use the 2 square identity....
Cheese3.1416
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Ok, well if\[130=11^2+3^2\]\[34=5^2+3^2\]then I guess\[(33+15)^2+(33-15)^2=4420\]
estudier
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Try again...
Cheese3.1416
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Eh, stupid me. I should've realized instantly that's wrong.
estudier
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Right idea, sort of...
estudier
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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)
Cheese3.1416
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I'm no good at being systematic.
estudier
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I will give you a medal for effort (and negative brownie points for being unsystematic)
Cheese3.1416
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That seems generous.
Cheese3.1416
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Damn. The moment I said this problem seemed recursive I should've realized I needed to do what you said 4 posts ago.
estudier
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Yes, procedure is simply a recursive algorithm when all is said and done...
Cheese3.1416
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Thanks for the medal, though I personally feel a bit disappointed with myself.
estudier
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S'OK, it's good practice...