Fermat Near Misses

I was surprised to find that my diurnal age today, 26914, is part of a triple of numbers that almost satisfies the equation \(x^3+y^3=z^3\) and hence the term "Fermat near misses". There are of course no positive integer solutions to this equation but some solutions only miss out by 1. This leads us to OEIS A050792:

A050792

Consider the Diophantine equation \(x^3 + y^3 = z^3 + 1  \, (1 < x < y < z) \) or 'Fermat near misses'. Arrange solutions by increasing values of \(z\) (see A050791). Sequence gives values of \(x\).

The initial \(x\) values are: 9, 64, 73, 135, 334, 244, 368, 1033, 1010, 577, 3097, 3753, 1126, 4083, 5856, 3987, 1945, 11161, 13294, 3088, 10876, 16617, 4609, 27238, 5700, 27784, 11767, 26914.

The corresponding \(z\) values are given by OEIS A050791:

A050791

Consider the Diophantine equation \(x^3 + y^3 = z^3 + 1 \, (1 < x < y < z)\) or 'Fermat near misses'. Sequence gives values of \(z\) in monotonic increasing order.

The initial \(z\) values are: 12, 103, 150, 249, 495, 738, 1544, 1852, 1988, 2316, 4184, 5262, 5640, 8657, 9791, 9953, 11682, 14258, 21279, 21630, 31615, 36620, 36888, 38599, 38823, 40362, 41485, 47584.

Given the \(x\) and \(z\) values, we can calculate the corresponding \(y\) values and so the initial \(x,y,z\) triples that satisfy \(x^3 + y^3 = z^3 + 1  \, (1 < x < y < z)\) are:

• (9, 10, 12)
• (64, 94, 103)
• (73, 144, 150)
• (135, 235, 249)
• (334, 438, 495)
• (244, 729, 738)
• (368, 1537, 1544)
• (1033, 1738, 1852)
• (1010, 1897, 1988)
• (577, 2304, 2316)
• (3097, 3518, 4184)
• (3753, 4528, 5262)
• (1126, 5625, 5640)
• (4083, 8343, 8657)
• (5856, 9036, 9791)
• (3987, 9735, 9953)
• (1945, 11664, 11682)
• (11161, 11468, 14258)
• (13294, 19386, 21279)
• (3088, 21609, 21630)
• (10876, 31180, 31615)
• (16617, 35442, 36620)
• (4609, 36864, 36888)
• (27238, 33412, 38599)
• (5700, 38782, 38823)
• (27784, 35385, 40362)
• (11767, 41167, 41485)
• (26914, 44521, 47584)

Specifically then we have \(26914^3+44521^3=47584^3+1\). The OEIS comments to A050792 are interesting:

Any number of solutions to the equation \(x^3 + y^3 = z^3 + 1\) can be produced through the use of the algebraic identity$$(9t^3+1)^3 + (9t^4)^3 = (9t^4+3t)^3 + 1$$by substituting in values of \(t\). Although these are certainly solutions, the identity generates only one family of solutions. Other solutions such as (64, 94, 103), (135, 235, 249) and (334, 438, 495) can be found. What is not known is if it is possible to parameterize all solutions for this equation. Put another way, are there an infinite number of families of solutions? Probable yes, but that too remains to be shown.

Here are the triples thrown up by the parameterized solutions for values of \(t\) from 1 to 20:

[9, 10, 12]
[73, 144, 150]
[244, 729, 738]
[577, 2304, 2316]
[1126, 5625, 5640]
[1945, 11664, 11682]
[3088, 21609, 21630]
[4609, 36864, 36888]
[6562, 59049, 59076]
[9001, 90000, 90030]
[11980, 131769, 131802]
[15553, 186624, 186660]
[19774, 257049, 257088]
[24697, 345744, 345786]
[30376, 455625, 455670]
[36865, 589824, 589872]
[44218, 751689, 751740]
[52489, 944784, 944838]
[61732, 1172889, 1172946] 
[72001, 1440000, 1440060]