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.