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sasogeek

  • 5 years ago

Show that there are no positive integers n for which n^4 + 2n^3 + 2n^2 + 2n + 1 is a perfect square. Are there any positive integers n for which n^4 +n^3 +n^2 +n+1 is a perfect square? If so, find all such n.

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  1. anonymous
    • 5 years ago
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    Hmm. This is tough. One idea is to try to force the number n^4 + 2n^3 + 2n^2 + 2n + 1 to be between two consecutive perfect squares. So we want to find some number N such that N^2 < n^4 + 2n^3 + 2n^2 + 2n + 1 < (N+1)^2. Looking at the problem, we might guess that N is a little bigger than n^2. If we let N = n^2 + C, where C is to be determined, then we've reduced the problem to finding a C such that (n^2 + C)^2 < n^4 + 2n^3 + 2n^2 + 2n + 1 < (n^2 + C + 1)^2. The second half of the inequality gives us reduces to n^4 + 2n^3 + 2n^2 + 2n + 1 < n^4 + c^2 + 1 + 2cn^2 + 2c + 2n^2. Cancel like terms, and see that we want to find a c such that 2n^3 + 2n < c^2 + 2cn^2 + 2c. This suggests maybe trying letting c = n. Now we want to see whether the first half of the long inequality is true. Well, it turns out that it is true. (n^2 + n)^2 < n^4 + 2n^3 + 2n^2 + 2n + 1 is a true statement. Therefore, n^4 + 2n^3 + 2n^2 + 2n + 1 can't be a perfect square because it's between (n^2 + n)^2 and (n^2 + n + 1)^2

  2. anonymous
    • 5 years ago
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    I can't figure out the second part :(. Wolfram alpha generated one solution n = 3, giving the square to be 121. I think that's probably the only solution. But I can't figure out why. I suspect you might use an argument which puts an upper and lower bound on the quantity again.

  3. sasogeek
    • 5 years ago
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    I'm not really sure. i found this from the archives of PROMYS

  4. anonymous
    • 5 years ago
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    What's PROMYS?

  5. sasogeek
    • 5 years ago
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    PROgram in Mathematics for Young Scientists

  6. anonymous
    • 5 years ago
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    Is that a high school program? Or college level program?

  7. sasogeek
    • 5 years ago
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    its kind of a college program organized for kids

  8. sasogeek
    • 5 years ago
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    http://math.bu.edu/people/promys/

  9. anonymous
    • 5 years ago
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    Ah, that's cool. I think I have a partial solution. If n is even, then let n = 2k. Then you can show that the number n^4 + n^3 + n^2 + n + 1 is bounded between ((2k)^2 + k)^2 and ((2k)^2 + k + 1)^2. Therefore, if n^4 + n^3 + n^2 + n + 1 is a square, n must be odd.

  10. anonymous
    • 5 years ago
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    For n = 2k + 1, we will try to bound the sum by (n^2 + k)^2 and (n^2 + k + 1)^2. Notice that k = (n-1)/2, that will be important. Anyways, (n^2 + k)^2 = n^4 + 2k n^2 + k^2. We'd like to show that's less than n^4 + n^3 + n^2 + n + 1. But n^4 + 2k n^2 + k^2 = n^4 + 2((n-1)/2) n^2 + k^2 = n^4 + n^3 - n^2 + k^2. Now we see that's less than n^4 + n^3 + n^2 + n + 1. Now consider (n^2 + k + 1)^2. We'd like to show that's bigger or equal to n^4 + n^3 +n^2 + n + 1. Well (n^2 + k + 1)^2 = n^4 + k^2 + 1 + 2kn^2 + 2k + 2n^2. Boil this down as much as you can, and use that k = (n-1)/2 for 2kn^2 again. Eventually, you want to show that k^2 + 2k bigger or equal to n, or k^2 + 2k is never less than 2k + 1. This is true, and in fact, equality occurs when k = 1. This equality for k = 1 gives us our unique solution n = 3. Phew.

  11. sasogeek
    • 5 years ago
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    thanks a lot :)

  12. sasogeek
    • 5 years ago
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    this is not my level of math but i'll try to go through over and over to get it :)

  13. anonymous
    • 5 years ago
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    No joke man. I'm a senior studying math in college and that was tough.

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