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.
Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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
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.
I'm not really sure. i found this from the archives of PROMYS
Not the answer you are looking for? Search for more explanations.
PROgram in Mathematics for Young Scientists
Is that a high school program? Or college level program?
its kind of a college program organized for kids
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.
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.
thanks a lot :)
this is not my level of math but i'll try to go through over and over to get it :)
No joke man. I'm a senior studying math in college and that was tough.