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- sasogeek

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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- sasogeek

- schrodinger

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- anonymous

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

- anonymous

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.

- sasogeek

I'm not really sure. i found this from the archives of PROMYS

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- anonymous

What's PROMYS?

- sasogeek

PROgram in Mathematics for Young Scientists

- anonymous

Is that a high school program? Or college level program?

- sasogeek

its kind of a college program organized for kids

- sasogeek

http://math.bu.edu/people/promys/

- anonymous

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.

- anonymous

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.

- sasogeek

thanks a lot :)

- sasogeek

this is not my level of math but i'll try to go through over and over to get it :)

- anonymous

No joke man. I'm a senior studying math in college and that was tough.

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