20060726, 15:29  #1 
Dec 2005
2^{2}×7^{2} 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 ? 
20060726, 15:59  #2 
Oct 2005
Fribourg, Switzerlan
374_{8} 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 20060726 at 16:42 
20060726, 16:36  #3  
Nov 2003
2^{2}×5×373 Posts 
Quote:
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.... 

20060726, 17:36  #4 
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 20060726 at 18:35 
20060726, 21:21  #5 
"Robert Gerbicz"
Oct 2005
Hungary
2^{2}·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. 
20060727, 12:40  #6  
Nov 2003
2^{2}×5×373 Posts 
Quote:
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. 

20060727, 13:29  #7  
Aug 2002
Buenos Aires, Argentina
2^{2}×3×113 Posts 
Quote:
Since a perfect square must be 0 or 1 (mod 8) we have: x^{2} = 0 or 1 (mod 8) x^{2} + 1 = 1 or 2 (mod 8) (1) r^{2} = 0 or 1 (mod 8) kr^{2} = 0 or k (mod 8) (2) Since x^{2} + 1 = kr^{2} 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 20060727 at 13:35 

20060727, 18:28  #8  
Jun 2003
2×5×7×71 Posts 
Quote:


20060727, 19:42  #9  
Aug 2002
Buenos Aires, Argentina
2^{2}×3×113 Posts 
Quote:
So we get: x^{2} = 0, 1, 4 (mod 8) x^{2}+1 = 1, 2, 5 (mod 8) kr^{2} = 0, k, 4k (mod 8) k = 1, 2, 5 (mod 8) r must still be odd. 

20060728, 01:21  #10 
Jan 2005
Transdniestr
1F7_{16} 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)?

20060728, 01:46  #11  
Aug 2002
Buenos Aires, Argentina
2^{2}×3×113 Posts 
Quote:
11*12*242*243 = 198^{3} 32*33*242*243 = 396^{3} 539*540*3024*3025 = 13860^{3} 

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