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Reducing to Pell's equation.

  1. May 5, 2004 #1


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    i am having some trouble with this problem.

    show that there are infinitely many integers n so that n^2+(n+1)^2 is a perfect square. (reduce to pell's equation).

    i know pell's equation but don't know how to apply it with this problem.

    pell's equation: n*x^2 + 1 = y^2.

  2. jcsd
  3. May 5, 2004 #2
    I've seen Pell's equation also written as [itex]nx^2 - 1 = y^2[/itex], so maybe there are two forms, one with a plus sign, one with a minus sign. One fact about Pell's equation is that there are an infinite number of positive integer pair solutions (x,y) when n is not a perfect square integer (hopefully this holds for the case where there's a minus sign, otherwise my hint will be useless!).

    Anyway, here's the hint: see what condition(s) you can extract when you try to solve the equation that results when you write [itex]n^2 + (n+1)^2[/itex] as a perfect square (i.e. when you solve the following equation for n: [itex]n^2 + (n+1)^2 = m^2[/itex])
  4. May 7, 2004 #3


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    hmmm, ok, but i still don't see how you can apply that to get further.
  5. May 8, 2004 #4
    Alright. So we want to solve the following equation for n:

    n^2 + (n+1)^2 = m^2

    where m is an integer. After a bit of manipulation we arrive at:

    2n^2 + 2n + (1 + m^2) = 0

    The solutions of this equation are:

    n = (-2 +/- sqrt(4 - 8(1 + m^2))/4)


    n = 1/2 (-1 +/- sqrt(2m^2 - 1))

    Now the only way that n is going to be an integer is if the expression inside the square root sign is a square number. Or in other words:

    2m^2 - 1 = k^2

    where k is an integer. This is the same as Pell's equation (with the minus sign), i.e.:

    Dx^2 - 1 = y^2

    Now we know that Pell's equation has infinitely many integer pair solutions (x,y) if D is not a square number. In our case, D is 2, which is not a square number. So there are an infinite number of integer pair solutions (m,k). And so there are an infinite number of integers n which are solutions to our first equation. And so there are infinitely many integers n such that n^2 + (n+1)^2 is a perfect square.
  6. May 30, 2004 #5
    In the Pell equation for a prime of the form p=4k+1 or 2, we have:

    Y^2 -pX^2 = +/- 1.

    Now for the Pathagorean triples, we have X^2 + Y^2 = Z^2, where it can be shown that absolute value of X=b^2-a^2, Absolute value of Y = 2ab, Z = a^2 + b^2, for a, b integers.

    Thus we have a form like N=b^2-a^2, N+1 = 2ab.

    Subtracting we get 1 = a^2 +2ab - b^2 = (a+b)^2 - 2b^2. This is a Pellian equation. Here we could also have the equation (b-a)^2 -2a^2 =-1, if N+1 had been chosen as absolute value of b^2-a^2.

    Take the case of 3^2 -2x2^2 = 1. In this case we have a+b =3, b =2, a=1,

    This gives: 3^2 +4^2 = 5^2. Now in the case, 7^2 -2x5^2 = -1. Here in the second case where n+1 = absolute value of b^2-a^2, we have b=2, -a=5. This gives the form 20^2 + 21^2 = 29^2.

    Since there is an infinite number of solutions to X^2-2Y^2 = +/- 1, we have an infinite number of solutions to n^2 + (n+1)^2 = u^2 in integers.
    Last edited: May 30, 2004
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