Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Proving a Polynomial is a Perfect Square

Date: 08/17/2007 at 13:47:51
From: Priyanka
Subject: proving that a polynomial is a perfect square.

Let a, b be odd integers such that a^2 - b^2 + 1 divides b^2 - 1. 
Prove that a^2 - b^2 + 1 is a perfect square.



Date: 08/21/2007 at 01:51:30
From: Doctor Vogler
Subject: Re: proving that a polynomial is a perfect square.

Hi Priyanka,

Thanks for writing to Dr. Math.  That's a really good question.  I've
been working on it since you posted it on Friday, because it really
intrigued me.  And I do love Diophantine equations.

A Diophantine equation is an equation (usually polynomials in at least
two variables) where you want solutions which are integers.  In your
case, the equation is

  (a^2 - b^2 + 1)k = b^2 - 1.

This equations has lots of solutions, infinitely many solutions.  And
you want to prove that all of them either have a or b even or a^2 -
b^2 + 1 a perfect square.  The thing that makes this especially hard
to prove is that when a or b is even, a^2 - b^2 + 1 is often not a
perfect square!  And there are infinitely many solutions where a or b
is even.

The first thing I noticed is that all of the solutions I found where a
and b were both odd had either a = b or had the form

  a = 4m^3 - m
  b = 4m^3 - 3m
  k = m^2 - 1

for some odd number m.

Later, I found solutions not of this form, but they were pretty large,
so I didn't find them until I understood this problem a lot better. 
So at first I wondered if all solutions with a and b both odd had this
form.  In particular, a^2 - b^2 + 1 being a square is equivalent to k
+ 1 being a square, since adding a^2 - b^2 + 1 to each side of your
original equation gives

  (a^2 - b^2 + 1)(k + 1) = a^2,

and a square times a non-square can never equal a square.  From the
same equation, it would suffice to prove that k + 1 was relatively
prime to (had no prime factors in common with) a^2 - b^2 + 1, which
seemed to be true, but I was unable to prove this.

It seemed better to consider it a kind of Pell Equation.  For example,
let's consider all integer solutions that have k = 1 (not just the odd
ones).  These have

  a^2 - b^2 + 1 = b^2 - 1

or

  a^2 = 2(b^2 - 1).

Since this implies that a^2 is even, that means that a is even, and

  2(a/2)^2 = b^2 - 1,

or

  b^2 - 2(a/2)^2 = 1.

This equation is familiar to someone who has studied number fields. 
It says that the norm of

  b + (a/2)*sqrt(2)

is 1.  It turns out that all such numbers are powers of 3 + 2*sqrt(2),
or the negative of such a power.  Since we don't care about the signs
of a and b, we can write

  b + (a/2)*sqrt(2) = (3 + 2*sqrt(2))^n

for some n.  For example, n = 3 gives (multiplying out the left side)

  (3 + 2*sqrt(2))^3 = 99 + 70*sqrt(2),

and so b = 99 and a/2 = 70, and

  (a, b, k) = (140, 99, 1)

is one solution to your equation.

In fact, we can generalize this.  You can always transform your
equation like so:

  (a^2 - b^2 + 1)k = b^2 - 1

  ka^2 = (k + 1)(b^2 - 1)

  k(k + 1)(a/(k + 1))^2 = b^2 - 1

  b^2 - (k(k + 1)) (a/(k + 1))^2 = 1.

Then you'll get a solution

  b + (a/(k + 1))*sqrt(k(k + 1))

for each power of

  (2k + 1) + 2*sqrt(k*(k + 1)).

For example, for the power n = 1, you get

  a = 2k + 2
  b = 2k + 1.

For n = 2, you get

  a = 8k^2 + 12k + 4
  b = 8k^2 + 8k + 1.

And if k = m^2 - 1 for some m, then you can transform your equation
like so

  (a^2 - b^2 + 1)(m^2 - 1) = b^2 - 1

  (m^2 - 1)a^2 = m^2(b^2 - 1)

  (m^2 - 1)(a/m)^2 = b^2 - 1

  b^2 - (m^2 - 1) (a/m)^2 = 1.

Then you'll get a solution

  b + (a/m)*sqrt(m^2 - 1)

for each power of

  m + sqrt(m^2 - 1).

For example, for the power n = 1, you get

  a = m
  b = m
  k = m^2 - 1.

For n = 3, you get

  a = 4m^3 - m
  b = 4m^3 - 3m
  k = m^2 - 1.

Importantly, if m is odd and n is also odd, then a and b are odd too.
It turns out that this will give you all solutions where a and b are
odd.  So let's get around to proving this:

Since we need to work with square roots of k(k + 1), we need to pull
out any square factors.  For example, if k = 3, then we need to work
with sqrt(3) instead of sqrt(12).  So let

  k(k + 1) = rs^2

where r is the square-free part of k(k + 1), and s is the largest
square that divides k(k + 1).  Since k(k + 1) can't be negative (even
if k is), we can assume that r and s are positive.  Then we transform
the equation into

  b^2 - r(ak/rs)^2 = 1.

So we this means that

  b + (ak/rs)*sqrt(r)

has norm 1.  Note that ak/rs is an integer; you see,

  (ka)^2 = k(k + 1)(b^2 - 1)

  (ka)^2 = rs^2(b^2 - 1),

which means that s^2 divides (ka)^2, and so s divides ka.  Also, r
divides (ka/s)^2, and since r is square-free, this implies that r
divides ka/s, so ka/rs is an integer.

By Dirichlet's Unit Theorem, all numbers of the form

  x + y*sqrt(r)

with x and y positive integers, which have norm 1, are powers of a
single such number.  That is, there are positive integers u and v
(depending only on k) such that

  x + y*sqrt(r) = (u + v*sqrt(r))^n

for some integer n.  (Actually, Dirichlet's Unit Theorem says more
than this, but since r is positive, it implies a fundamental unit in
the ring of integers of Q(sqrt(r)), and all numbers with norm 1 will
either be powers of this unit or powers of its square, and all numbers
with x and y integers will either be powers of this unit (or its
square) or powers of the cube.  So the number u + v*sqrt(r) is the
fundamental unit raised to the 1, 2, 3, or 6.)

Now, we have one other important fact.  The number

  (2k + 1) + (2s)*sqrt(r)

also has norm 1, since

  (2k + 1)^2 - r(2s)^2 = (4k^2 + 4k + 1) - 4(k^2 + k) = 1,

so it is also a power of u + v*sqrt(r).

Now consider what is the largest power of 2 which divides v.  Say that
it is t, so that v/2^t is an odd integer.  Then a little modular
arithmetic will show that odd powers of u + v*sqrt(r) will have the form

  x + y*sqrt(r)

where y is divisible by 2^t but not 2^(t+1), but even powers of u +
v*sqrt(r) will have y divisible by 2^(t+1).  Now, we know that

  (2k + 1) + (2s)*sqrt(r)

is such a power, so the question is whether 2^(t+1) divides 2s.  If it
does, then that means that  

  (2k + 1) + (2s)*sqrt(r)

is an even power of u + v*sqrt(r), so it is the square of some other
such power, say

  (2k + 1) + (2s)*sqrt(r) = (x + y*sqrt(r))^2
                          = (x^2 + ry^2) + (2xy)*sqrt(r)

where

  x^2 - ry^2 = 1,

but then

  2k + 1 = x^2 + ry^2 = 2x^2 - 1,

and so

  k + 1 = x^2

is a perfect square, as we needed to prove.

The alternative is 2^(t+1) does not divide 2s, but only 2^t.  But then
that means that

  b + (ak/rs)*sqrt(r),

which is also a power of u + v*sqrt(r), must have

  2^t divides (ak/rs).

But

  ak/rs = as/(k + 1),

so

  2^t divides as/(k + 1).

But if a and b are both odd, then k is even, and so k + 1 is odd. 
Therefore,

  2^t divides as.

But since 2^(t+1) does not divide 2s, that means that 2^t does not
divide s, and therefore a must be even.

If you have any questions about this or need more help, please write
back and show me what you have been able to do, and I will try to
offer further suggestions.

- Doctor Vogler, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Number Theory

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/