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Topic: Divide two expressions, get perfect square? (Read 1727 times) |
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K Sengupta
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Divide two expressions, get perfect square?
« on: Jul 27th, 2007, 8:06am » |
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Analytically determine whether there exists any pair of positive integers (x, y) satisfying:
(x2000 – 1)/(x-1) = y2
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Eigenray
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Re: Divide two expressions, get perfect square?
« Reply #1 on: Jul 31st, 2007, 10:58am » |
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I must be missing the "easy" solution! First observe:
| hidden: | x2000-1 =   n(x),
where the product is over divisors n of 2000, and n is the n-th cyclotomic polynomial.
Now, for n  m, n and m are relatively prime, since they are both irreducible, and therefore 1 is a linear combination of n and m, with coefficients in  [x]. Clearing denominators, we get a relation of the form d = f n + g m, where d is an integer, and f,g   [x]. Then for any integer x, gcd(n(x), m(x)) must divide d. For any given n,m, this value d can be found using the Euclidean algorithm, and it turns out that for n,m divisors of 2000, d is either 1,2, or 5. |
Here's a question (but this goes beyond easy, I'm sure): find a closed form for d, in terms of n and m.
Anyway, given the above, lets find all integer solutions:| hidden: |
Since any two factors n(x) and m(x) have a gcd of 1,2, or 5, and the product is a square, it follows that each factor is of the form 2a5b*y2, where a,b are either 0 or 1. [Note that for n>2, n(x)>0 for all x, since it has no real root.]
Since 8(x) = x4+1 is never divisible by 5, it is either a square, or twice a square. But if x4+1=y2, then 1=y2-x4 is the difference of two squares, which happens only for y2=1, x=0. Aside from this solution, we can conclude therefore that x4+1 is twice a square, and therefore x is odd.
Now observe that 5(x) = 1+x+x2+x3+x4, and 10(x) = 5(-x) are always odd. Moreover, 5(x) is divisible by 5 iff x=1 mod 5, and 5 | 10(x) iff x=-1 mod 5. WLOG, then (replacing x with -x if necessary), 5(x) is not divisible by 5.
It follows now that 5(x) must be a square. But! x is odd, and
(x2+(x-1)/2)2 < x4+x3+x2+x+1 < (x2+(x+1)/2)2,
since the difference on the left is 7x2/4+3x/2+3/4 > 0, and the difference on the right is x2/4 - x/2 - 3/4 > 0 when x < -1, or x > 3. This leaves only x=-1,1,3 to consider, which is straightforward. |
Thus the only integer solutions are (0, 1), and (-1, 0). Phew!
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| « Last Edit: Aug 5th, 2007, 12:20am by Eigenray » |
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towr
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Re: Divide two expressions, get perfect square?
« Reply #2 on: Jul 31st, 2007, 12:16pm » |
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Can't we do something with (x2000 – 1)/(x-1) =  1999i=0 xi?
[e] If I'm not mistaken xi mod m is periodic with period m or some smaller factor of it. So among other things (x2000 – 1)/(x-1) should be 0 modulo every m that divides 2000. Right? But is that remotely useful... hmm[/e]
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| « Last Edit: Jul 31st, 2007, 12:39pm by towr » |
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Eigenray
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Re: Divide two expressions, get perfect square?
« Reply #3 on: Jul 31st, 2007, 3:40pm » |
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I'm not sure what.
If 2000 were odd, say f(x) is a monic polynomial of even degree 2n, we could expand  f(x) = g(x) + O(1/x), where g is a polynomial with rational coefficients of degree n. If f is not a perfect square, then f(x) - g(x)2 = c*xk + O(xk-1), with c 0, and k < n.
For each congruence class of x mod the LCM of the denominators of g's coefficients, find a rational t, 0  t 1, such that g(x)-t is integer (don't take t=1 if c>0, nor t=0 if c<0). Then
(g(x)-t)2 < f(x) < (g(x)+1-t)2
for x sufficiently large. More generally, I guess if f(x) is monic of degree kn, and is not a perfect k-th power, then f(x)=yk has only finitely many solutions, which can be found constructively? (Sounds right but if so I'm surprised I didn't know that.)
I didn't get anywhere considering (x2000-1)/(x-1) mod m for various m.
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K Sengupta
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Re: Divide two expressions, get perfect square?
« Reply #4 on: Aug 6th, 2007, 8:05am » |
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A proposed solution is furnished hereunder as follows:
Substituting (a, b, c) = {x^1000 + 1, x^500 + 1, (x^500 – 1)/(x-1)}
we obtain:
(x^2000 – 1)/(x-1) = abc
Now b and c divide a-2, while c divides b-2.
Accordingly, it follows that the gcd of any two of a, b and c is at most 2.
The product abc is a perfect square if a, b and c are either squares or doubles of squares.
It is clearly observed that neither a nor b can be squares of positive integers.
Accordingly, each of a and b must correspond to twice the square of positive integers.
Therefore, we must have:
4ab = 4*x^1500 + 4*x^1000 + 4*x^500 + 4 must be a perfect square.
However, in that situation:
(2*x^750 + x^250)^2 < 4*x^1500 + 4*x^1000 + 4*x^500 + 4
< (2*x^750 + x^250 + 1)^2
This implies that 4ab is not a perfect square, which leads to a contradiction.
Consequently, the given equation does not admit of positive integer solutions.
Of course, the non negative integer solutions are indeed : (0, +/-1), and (-1, 0) as rightly pointed out by Eigenray.
In conclusion, I personally consider the above solution in terms of the current post to be much inferior in quality when compared to Eigenray's superior methodology.
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| « Last Edit: Aug 6th, 2007, 9:37am by K Sengupta » |
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Eigenray
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Re: Divide two expressions, get perfect square?
« Reply #5 on: Aug 7th, 2007, 6:21am » |
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on Aug 6th, 2007, 8:05am, K Sengupta wrote:| In conclusion, I personally consider the above solution in terms of the current post to be much inferior in quality when compared to Eigenray's superior methodology. |
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 I don't know why you would say that. Yours is more elementary and only requires that 2000 = 8n. Mine only works for 2000 = 40n, where n is a product of powers of 2 and 5.
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