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anonymous

  • 5 years ago

The sum of the squares of two consecutive odd integers is 74. Find the two integers.

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  1. dumbcow
    • 5 years ago
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    odd integer can be represented as 2n+1 and the next odd int would 2n+3 so you have: (2n+1)^2 + (2n+3)^2 = 74 solve for n

  2. anonymous
    • 5 years ago
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    i dont understand it..

  3. anonymous
    • 5 years ago
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    i dont remember how to solve n

  4. dumbcow
    • 5 years ago
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    hmm i guess an easier way is to look at perfect squares and which ones add up to 74 1,3 = 1^2 + 3^2 = 10 3,5 = 3^2 + 5^2 = 34 ....

  5. anonymous
    • 5 years ago
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    so its 3 and 5?

  6. dumbcow
    • 5 years ago
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    just remember when you find it that their negative counterparts are answers too because we square them -5,-3, = (-5)^2 + (-3)^2 = 34 ...

  7. dumbcow
    • 5 years ago
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    no sum of 3 and 5 squared is 34...keep going try 5,7 7,9 9,11 has to equal 74

  8. anonymous
    • 5 years ago
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    5 and 7

  9. dumbcow
    • 5 years ago
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    correct

  10. anonymous
    • 5 years ago
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    thank you!

  11. dumbcow
    • 5 years ago
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    your welcome

  12. anonymous
    • 5 years ago
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    Your first method was nicer :( You can't brute force it once you get up to bigboy numbers

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