Find the smaller of two consecutive positive 30-digit integers such that the difference of their cubes is a perfect square.
That is: (j+1)3- j3 = k2, where j and k are positive integers.
That is: (j+1)3- j3 = k2, where j and k are positive integers.
We are to find j such that j has 30 digits.
The sequence starts, {0,7,104,1455,20272,...}
My normal approach with such problems is to take the multiple between consecutive terms, which is 7+4sqrt(3) or around 13.9282 for each successive term, and then apply this multiple to compute the integer part of the next term. This works because the conjugate pair of the difference, (7-4sqrt(3))^n becomes small enough to be rounded safely.
However, in this case a 30 digit number is required, while excel becomes imprecise long before that.
So in this case an alternative line of attack is needed. We can also use the recurrence a(n)=14a(n-1)-a(n-2)+6. which requires only addition and subtraction.
Then we obtain {...282359,3932670,54776287,...} and after some repetitive arithmetic:
296,892,618,846,717,316,677,163,422,344
which is the required value of j, when k=71233×7219029104975160489230437
Paradoxically, finding a 30-digit solution is much easier than one with 10 or 18 digits.
Edited on December 13, 2024, 12:23 am
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Posted by broll
on 2024-12-13 00:03:26 |
Possible Solution
