Number Theory - Perfect Square
Date: 5/26/96 at 16:17:12 From: Anonymous Subject: Number Theory I have been unable to solve this problem: "Determine all the possible values of n for which the number n^4 - 2*n^3 + 4*n^2 - 6*n + 3 is the square of an entire number (i.e belongs to N)." I think I have to group this in products of polinomy, so it becomes: ((n - 1)^2) * (n^2 + 3) In this way we can note that the first part is a square, for each value of n, while the second cannot be a square. The only number is 1, but in this way the expression become 0 So no entire number exists. Is this resolution right?
Date: 5/27/96 at 21:46:31 From: Doctor Ken Subject: Re: Number Theory Hello - Yes, your solution is correct. The only perfect square ("perfect square" is our name for your "square of an entire number") that is 3 more than another perfect square is 4, so n has to be 1. The other thing that you're using in this problem is that if you put an integer in prime-factored form, then the power on each prime must be an even number. And all the powers from the (n-1)^2 part will be even, so all the powers on the (n^2 + 3) part have to be even too - thus (n^2 + 3) is a perfect square. Good factoring! -Doctor Ken, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 03/20/99 at 05:38:22 From: Anandh Pachaiappan Subject: Number Theory Hello, I was going through your archives, and I have one suggestion for the problem Number Theory - Perfect Square: Determine all the possible values of n for which the number n^4 - 2*n^3 + 4*n^2 - 6*n + 3 is the square of an entire number (i.e belongs to N). It is factored as ((n - 1)^2) * (n^2 + 3), and you suggested only n = 1 satisfies which gives the result as 0. Actually n= -1 is also another solution for which we get N = 16 which is a perfect square. This is because we want n^2 + 3 to be a perfect square. Then the number will factor into: (n - 1) * sqrt(n^2 + 3) and will be a whole number if n^2 + 3 is a perfect square. As Doctor Ken states, the only perfect square that is 3 more than another perfect square is 4. So we need: n^2 + 3 = 4 n^2 = 1 n = +/- 1 Then we get 0 and 16, which correspond to the values n = 1 and n = -1. Regards, P.Anandh.
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