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estudier Group Title

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

  • one year ago
  • one year ago

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  1. Cheese3.1416 Group Title
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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.

    • one year ago
  2. estudier Group Title
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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

    • one year ago
  3. estudier Group Title
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    aka the Brahmagupta–Fibonacci identity

    • one year ago
  4. Cheese3.1416 Group Title
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    You seem to know what you're doing better than I do. Is this a test?

    • one year ago
  5. estudier Group Title
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    Think of it as a tutorial...

    • one year ago
  6. Cheese3.1416 Group Title
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    OK, challenge accepted. Whether it'll be completed is entirely another matter....

    • one year ago
  7. estudier Group Title
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    :-)

    • one year ago
  8. Cheese3.1416 Group Title
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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.

    • one year ago
  9. shubhamsrg Group Title
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    *

    • one year ago
  10. estudier Group Title
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    OK, a hint then: 5 is a factor of 4420 and you could write that as (2^2 +1^2)

    • one year ago
  11. Cheese3.1416 Group Title
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    So now the problem becomes write 884 as the sum of two squares?

    • one year ago
  12. estudier Group Title
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    Well, yes, 5 is not the only factor....

    • one year ago
  13. Cheese3.1416 Group Title
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    This is starting to look recursive...

    • one year ago
  14. estudier Group Title
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    First off, you better get a factorization of 4420

    • one year ago
  15. Cheese3.1416 Group Title
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    \[2 \times 2 \times 5 \times 221\]

    • one year ago
  16. Cheese3.1416 Group Title
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    \[221=14^2+5^2\]

    • one year ago
  17. estudier Group Title
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    Yes, u could have bro0ken 221 into 13 *17

    • one year ago
  18. Cheese3.1416 Group Title
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    \[20=2^2+4^2\]

    • one year ago
  19. estudier Group Title
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    There is more than 1 answer so let's use 2^2 * 5 * 13 * 17 as the factorization

    • one year ago
  20. Cheese3.1416 Group Title
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    I really need to make myself more intelligent.

    • one year ago
  21. estudier Group Title
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    That is just the usual factorization...

    • one year ago
  22. estudier Group Title
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    Small numbers , easy to put as sums of squares...

    • one year ago
  23. Cheese3.1416 Group Title
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    I am now processing.

    • one year ago
  24. estudier Group Title
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    Save a little time (2^2+0^2) (2^2+1^2) (3^2+2^2) (4^2+1^2)

    • one year ago
  25. Cheese3.1416 Group Title
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    There was no need to give me the answer.

    • one year ago
  26. Cheese3.1416 Group Title
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    You spoiler.

    • one year ago
  27. estudier Group Title
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    That's not the answer....

    • one year ago
  28. Cheese3.1416 Group Title
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    Oh.

    • one year ago
  29. Cheese3.1416 Group Title
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    Well I can work it out myself.

    • one year ago
  30. Cheese3.1416 Group Title
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    I'm about to check 130 and 34.

    • one year ago
  31. estudier Group Title
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    But u don't have to check anything, just use the 2 square identity....

    • one year ago
  32. Cheese3.1416 Group Title
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    Ok, well if\[130=11^2+3^2\]\[34=5^2+3^2\]then I guess\[(33+15)^2+(33-15)^2=4420\]

    • one year ago
  33. estudier Group Title
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    Try again...

    • one year ago
  34. Cheese3.1416 Group Title
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    Eh, stupid me. I should've realized instantly that's wrong.

    • one year ago
  35. estudier Group Title
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    Right idea, sort of...

    • one year ago
  36. estudier Group Title
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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)

    • one year ago
  37. Cheese3.1416 Group Title
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    I'm no good at being systematic.

    • one year ago
  38. estudier Group Title
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    I will give you a medal for effort (and negative brownie points for being unsystematic)

    • one year ago
  39. Cheese3.1416 Group Title
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    That seems generous.

    • one year ago
  40. Cheese3.1416 Group Title
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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.

    • one year ago
  41. estudier Group Title
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    Yes, procedure is simply a recursive algorithm when all is said and done...

    • one year ago
  42. Cheese3.1416 Group Title
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    Thanks for the medal, though I personally feel a bit disappointed with myself.

    • one year ago
  43. estudier Group Title
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    S'OK, it's good practice...

    • one year ago
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