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Old 2006-07-26, 15:29   #1
Kees
 
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Default On consecutive integers

Continuing on the problem of four consecutive integers, can we find two pairs of positive consecutive integers such that there product is a square and all four integers are different ?
And can we find more than one solution ?
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Old 2006-07-26, 15:59   #2
victor
 
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First pair is p1, second one is p2 :

With p1 = {0;1} or {-1;0}
any p2 suits to the problem.
So there is an infinity of solutions.

[EDIT] Sorry, I had not seen the "positive". :(
I first saw "four consecutive integers" and did not pay attention to "two pairs of _positive_ consecutive integers".
Mea culpa.

Last fiddled with by victor on 2006-07-26 at 16:42
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Old 2006-07-26, 16:36   #3
R.D. Silverman
 
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Quote:
Originally Posted by victor
First pair is p1, second one is p2 :

With p1 = {0;1} or {-1;0}
any p2 suits to the problem.
So there is an infinity of solutions.
Uh.... the original post said "positive integers".

One possibility: consider the following pairs:

(a, a+1) and (a(a+1) , a^2+a+1) such that a^2 + a + 1 is a square.

Whence:

y^2 = a^2 + a + 1 --> y^2 = (a + 1/2)^2 + 3 --> (2y)^2 = (2a+1)^2 + 3

The only squares that differ by 3 are 1 and 4, giving y = 1 and a = 0;
Thus there are no solutions of this form.

Interesting problem. I suspect there are no solutions; Let me think
about it....
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Old 2006-07-26, 17:36   #4
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Default One low solution

1*2*8*9 = 12^2

[EDIT: Some more]
3*4*48*49 = 168^2
5*6*120*121 = 660^2

There's 21 solutions where all four integers are < 1000 and at least 76 where all 4 integers are < 10000.

The biggest one I found was 1681*1682*9800*9801 (but this was just a couple of minutes of searching).

===================================================

I will try searching for pairs of consecutive triplets now.

Last fiddled with by grandpascorpion on 2006-07-26 at 18:35
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Old 2006-07-26, 21:21   #5
R. Gerbicz
 
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There are infinite many solutions.
Proof: there are infinite many solutions of the x^2=2*y^2+1 Pell equation. Let (x1,y1) and (x2,y2) two different solutions of the equation and let a=2*y1^2 and b=2*y2^2. Then a isn't equal to b and
a*(a+1)*b*(b+1)=2*y1^2*x1^2*2*y2^2*x2^2=(2*x1*y1*x2*y2)^2 so the product is square, what was required.
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Old 2006-07-27, 12:40   #6
R.D. Silverman
 
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Quote:
Originally Posted by R. Gerbicz
There are infinite many solutions.
Proof: there are infinite many solutions of the x^2=2*y^2+1 Pell equation. Let (x1,y1) and (x2,y2) two different solutions of the equation and let a=2*y1^2 and b=2*y2^2. Then a isn't equal to b and
a*(a+1)*b*(b+1)=2*y1^2*x1^2*2*y2^2*x2^2=(2*x1*y1*x2*y2)^2 so the product is square, what was required.
Another method: Find a square x^2 such that x^2 + 1 = kr^2 for some
arbitrary k. This is another Pell equation. (with -1 instead of +1)

Now let the second pair be (ka^2, ka^2 + 1) such that ka^2 + 1 is
also a square. This is the same Pell equation as above with +1 instead of -1.


The above post took k = 2. This generalizes it.
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Old 2006-07-27, 13:29   #7
alpertron
 
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Quote:
Originally Posted by R.D. Silverman
Another method: Find a square x^2 such that x^2 + 1 = kr^2 for some
arbitrary k. This is another Pell equation. (with -1 instead of +1)

Now let the second pair be (ka^2, ka^2 + 1) such that ka^2 + 1 is
also a square. This is the same Pell equation as above with +1 instead of -1.


The above post took k = 2. This generalizes it.
Not all values of k can be used in that Pell equation.

Since a perfect square must be 0 or 1 (mod 8) we have:

x2 = 0 or 1 (mod 8)
x2 + 1 = 1 or 2 (mod 8) (1)

r2 = 0 or 1 (mod 8)
kr2 = 0 or k (mod 8) (2)

Since x2 + 1 = kr2 from (1) and (2) we get that k=1 or 2 (mod 8). We can also say that r must be odd.

Last fiddled with by alpertron on 2006-07-27 at 13:35
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Old 2006-07-27, 18:28   #8
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Quote:
Originally Posted by alpertron
Since a perfect square must be 0 or 1 (mod 8) we have:
A perfect square must be 0,1 or 4 (mod 8), no?
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Old 2006-07-27, 19:42   #9
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Quote:
Originally Posted by axn1
A perfect square must be 0,1 or 4 (mod 8), no?
Yes, you're right.

So we get:

x2 = 0, 1, 4 (mod 8)
x2+1 = 1, 2, 5 (mod 8)

kr2 = 0, k, 4k (mod 8)

k = 1, 2, 5 (mod 8)

r must still be odd.
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Old 2006-07-28, 01:21   #10
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Default Cubes?

Now that squares are out of the way ...Are there any cases where the product of two different pairs of positive consecutive numbers is a cube (of an integer)?
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Old 2006-07-28, 01:46   #11
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Quote:
Originally Posted by grandpascorpion
Now that squares are out of the way ...Are there any cases where the product of two different pairs of positive consecutive numbers is a cube (of an integer)?
I found three small instances:

11*12*242*243 = 1983
32*33*242*243 = 3963
539*540*3024*3025 = 138603
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