
COMMENTS

That is, numbers n >= 0 such that 5n(n^2+6) is a square.
According to Dickson, Lucas stated that the only terms are 2, 3, 98 and 120 (missing 27).
The Mordell reference shows that there are only finitely many solutions.  Allan Wilks, Mar 10 2007
It can be shown that all such numbers n can be obtained from elements that are perfect squares in the following 3 recurrent sequence:
1) x(0)=0, x(1)=4, x(k+1) = 98*x(k)  x(k1). If x(k) is a square then n = 30*x(k). In particular: for k=0, we have n=30*x(0)=0, for k=1, we have n=30*x(1)=120.
2) x(0)=1, x(1)=49, x(k+1) = 38*x(k)  x(k1). If x(k) is a square then n = 2*x(k). In particular: for k=1, we have n=2*x(1)=2, for k=2, we have n=2*x(1)=98.
3) x(0)=1, x(1)=9, x(k+1) = 8*x(k)  x(k1). If x(k) is a square then n = 3*x(k). In particular: for k=1, we have n=3*x(1)=3, for k=2, we have n=3*x(1)=27.
It also follows that for any such n one of n/2, n/3, or n/30 is a perfect square. I have tested 10^5 terms of each of the recurrent sequences above and found no new perfect squares. (End)
The sequence is finite because the number of integral points on an elliptic curve is finite; in this case the curve is m^2 = 5n^3 + 30n. Multiplying the equation by 25 and letting y = 5m and x = 5n, we have y^2 = x^3 + 150x. According to Magma, the integral points on this curve are (x, y) = (0, 0), (10, 50), (15, 75), (24, 132), (135, 1575), (490, 10850), (600, 14700). So the list is complete. (End)
This was also confirmed (using Sage) by Jaap Spies, May 27 2007
About the second comment: Lucas, however, did actually include 27 in his original note, which can be seen at the link cited. The mistake appears to have originated with Dickson.  Matt Westwood, Mar 05 2022


REFERENCES

L. E. Dickson, History of the Theory of Numbers, Volume 2, Chapter 21, page 587.
L. J. Mordell, Diophantine Equations, Ac. Press; see Th. 1, Chap. 27, p. 255.
