Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Reducing to Pell's equation.

  1. May 5, 2004 #1


    User Avatar

    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


    User Avatar

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Reducing to Pell's equation.
  1. Pell's Equation (Replies: 4)