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 2006-07-26, 15:29 #1 Kees     Dec 2005 22×72 Posts 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 ?
 2006-07-26, 15:59 #2 victor     Oct 2005 Fribourg, Switzerlan 3748 Posts 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
2006-07-26, 16:36   #3
R.D. Silverman

Nov 2003

22×5×373 Posts

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

 2006-07-26, 17:36 #4 grandpascorpion     Jan 2005 Transdniestr 503 Posts 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
 2006-07-26, 21:21 #5 R. Gerbicz     "Robert Gerbicz" Oct 2005 Hungary 22·367 Posts 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.
2006-07-27, 12:40   #6
R.D. Silverman

Nov 2003

22×5×373 Posts

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.

2006-07-27, 13:29   #7
alpertron

Aug 2002
Buenos Aires, Argentina

22×3×113 Posts

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

2006-07-27, 18:28   #8
axn

Jun 2003

2×5×7×71 Posts

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?

2006-07-27, 19:42   #9
alpertron

Aug 2002
Buenos Aires, Argentina

22×3×113 Posts

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.

 2006-07-28, 01:21 #10 grandpascorpion     Jan 2005 Transdniestr 1F716 Posts 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)?
2006-07-28, 01:46   #11
alpertron

Aug 2002
Buenos Aires, Argentina

22×3×113 Posts

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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