- #1

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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.

-thanks.

- Thread starter qaz
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- #1

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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.

-thanks.

- #2

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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])

- #3

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cragwolf said:

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])

hmmm, ok, but i still don't see how you can apply that to get further.

- #4

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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)

Or:

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.

- #5

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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.

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.

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