Find all n^3-n = d^2+d

  • Thread starter ramsey2879
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Someone told me that he heard that there is a proof based on the elliptic method that only one solution in intergers exists for this problem. This is beyond me however as I have no experience in this area. Can anyone lead me to a citation of the proof or tell me how to proceed. P.S. I determined that d has to equal the floor of the square root of n^3-n and that n=6, d=14 is a solution.
 
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Zurtex
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n^3 - n = d^2 + d

n(n^2 - 1)= d(d + 1)
n(n + 1)(n - 1) = d(d + 1)

Clearly n = 2 and d = 2 is a solution. n = 0 and d = 0 or -1, n = -1 and d = 0 or -1, n = 1 and d = 0 or -1...
 
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A start on the problem is to notice that 4(d(d+1))+1 = 4d^2+4d+1 = (2d+1)^2.

Thus we have 4N^3-4N+1 = X^2.

Checking for squares up to N=1000, N non-negative, produces only N=0,1,2,6. This suggests that no other solutions exist
 
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robert Ihnot said:
A start on the problem is to notice that 4(d(d+1))+1 = 4d^2+4d+1 = (2d+1)^2.

Thus we have 4N^3-4N+1 = X^2.

Checking for squares up to N=1000, N non-negative, produces only N=0,1,2,6. This suggests that no other solutions exist
Yes, you are right I miss some important conditions given in the original problem thst excluded n=0,1,2. In my modification I let n=b in the problem below: Let a, b, c,d,e be integers such that none of them equals 1.
Does there exist a set of three consecutive integers,{a,b,c} and
another set of two consecutive integers, {d,e} such that a*b*c = d*e?
So n=0,1,2 are excluded since then one of a,b,c=1. But then N=-1 appears to be a solution, doesn't it?
 

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