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.
(j+1)^3 - j^3 = 3j^2 + 3j + 1
call f(j) = 3j^2 + 3j + 1
Checking j values that result in f(j) being a perfect square gives a sequence which is in Sloanes: A001921
[0, 7, 104, 1455, 20272, 282359]
a(n) = 14*a(n-1) - a(n-2) + 6 for n>1, a(0)=0, a(1)=7.
A program to compute the subsequent values and print any that are 30 digits long finds:
296892618846717316677163422344
(And since the next member of the sequence is almost 14 times the previous member, there are never two or more members with the same number of digits, so this is the only 30 digit solution)
For this value of j, the value of f(j) is:
264435681376986499276099298725117359547107705855947045630041
which is 60 digits long
Using full precision calculator to find its square root shows:
514233100234695607129351718821
(which, when squared, equals the 60 digit number, so "check").
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Posted by Larry
on 2024-12-13 14:13:02 |
solution
