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 1,3,8 ...extend! (Posted on 2017-01-26)
The set (1,3,8) has this peculiar feature:

Multiply any two of these numbers together and add 1 and you'll always get a perfect square:

1+ 1 × 3 = 4
1 + 1 × 8 = 9
1 + 3 × 8 = 25

a. Add to the above set a 4th integer, thus enabling 3 additional equalities.
b. Can you show why no 5th member maintaining the above feature exists?

 No Solution Yet Submitted by Ady TZIDON Rating: 5.0000 (1 votes)

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 results from some reading | Comment 2 of 4 |
This is a very old problem, dating to Diophantus of Alexandria in the 3rd century, who first observed that the quadruple of rational numbers {1/16, 33/16, 17/4, 105/16} has the property that the product of any two of them plus one is the square of a rational number.

In the 17th century Pierre de Fermat found the integer solution {1,3,8,120}.

Leonhard Euler extended Fermat's solution using the rational number 777480/8288641.  Euler also found an infinite family of such sets:
{a, b, a + b + 2r, 4r(r + a)(r + b) }, where ab + 1 = r^2.

In 1969 Baker and Davenport used 'Baker's theory on linear forms in logarithms of algebraic numbers and a reduction method based on continued fractions' to prove if d is a positive integer such that {1, 3, 8, d} forms a Diophantine quadruple, then d = 120.  By implication, Fermat's set cannot be extended to a Diophantine quintuple.

In 1979 a constructive proof was found to extend any Diophantine triple to a Diophantine quadruple.

In 2004 Dujella proved no Diophantine 6-tuple exists.  That capped the upward search since any 6 terms of a 7-tuple would be a 6-tuple.

No Diophantine quintuple has been found nor has one been proved impossible.

https://web.math.pmf.unizg.hr/~duje/intro.html

 Posted by xdog on 2017-01-26 17:03:43

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