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Cheese3.1416
 2 years ago
Best ResponseYou've already chosen the best response.1I 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
 2 years ago
Best 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

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1aka the Brahmagupta–Fibonacci identity

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

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1Think of it as a tutorial...

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

Cheese3.1416
 2 years ago
Best ResponseYou've already chosen the best response.1Maybe this is me being stupid, but all I can say about the two square identity is it gives me one equation with 4 unknowns.

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

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

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1Well, yes, 5 is not the only factor....

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

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1First off, you better get a factorization of 4420

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

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1Yes, u could have bro0ken 221 into 13 *17

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

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

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1That is just the usual factorization...

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1Small numbers , easy to put as sums of squares...

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

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1Save a little time (2^2+0^2) (2^2+1^2) (3^2+2^2) (4^2+1^2)

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

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1That's not the answer....

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

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

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

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

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

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1Right idea, sort of...

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1More 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
 2 years ago
Best ResponseYou've already chosen the best response.1I'm no good at being systematic.

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1I will give you a medal for effort (and negative brownie points for being unsystematic)

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

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

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1Yes, procedure is simply a recursive algorithm when all is said and done...

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

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1S'OK, it's good practice...
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