The Congruent Number 5 Problem  

Michael Somos described this problem to me; it is a problem he worked on early in his career.

Find a two rational numbers a and b such that a2-5b2 and a2+5b2 are both squares of rational numbers.

Any solution can be converted to an integer solution by multiplying all numbers by the LCM of the denominators of a and b.

History

This problem has quite a history going back to classical Greece; it is a specific class of solutions to Diophantine equations involving squares in arithmetic progression. I'll note that for integer a and k, if a2+k and a2-k are both squares, then the valid values k are: 24, 96, 120, 216, 240, 336, 384, ... (OEIS sequence A256418).

Fibonacci wrote a not-very-well-known book The Book of Squares addressing this more general problem, which is called the congruum problem. Example solutions include:

292 - 840 = 1 = 12 ; 292 + 840 = 1681 = 412
372 - 840 = 529 = 232 ; 372 + 840 = 2209 = 472

Sequences with the Generating Function p(t)/(1-ct+t2)

The even-indexed Fibonacci numbers F2n = 0, 1, 3, 8, 21, 55, 144, 377, 987, 2584, 6765, ... (Sloane's A1906) have generating function

g.f.: *t/(1-3t+t2)

and so are defined by the recurrence

A0 = 0; A1 = 1; An = 3An-1 - An-2

The Chebyshev_polynomials have the generating functions and recurrence definitions

g.f.: 1-*xt/(1-2xt+t2)
T0 = 1; T1 = x; Tn = 2xTn-1 - Tn-2

and

g.f.: 1/(1-2*xt+t2)
U0 = 1; U1 = 2x; Un = 2xUn-1 - Un-2

Michael Somos, "In the Elliptic Realm", writes:

[...] the following equations

    s(2n+2)*s(1) = s(n+2)*s(n+1) - s(n+1)*s(n) , and
    s(2n+1)*s(1) = s(n+1)*s(n+1) - s(n)*s(n) ,

hold for all positive n. These equations can be solved for s(n) where n is greater than 2 given any values for s(2) and s(1) but s(1) must be non-zero. When s(1) = 1 and s(2) = x , these sequences are related to Chebyshev polynomials of the second kind as follows

    s(n) = U(n-1, x/2).

Since "s(1)" is 1, the definitions are equivalent to:

s(1) = 1
s(2) = x
s(2n+1) = s(n+1)2 - s(n)2
s(2n+2) = s(n+2)s(n+1) - s(n+1)s(n)

Letting n=1, we get:

s(3) = x2-1
s(4) = s(3)x - x = x3 - 2x

Letting n=2, we get:

s(5) = s(3)2 - s(2)2 = x4-2x2+1 - x2 = x4 - 3x2 + 1
s(6) = s(4)s(3) - s(3)s(2) = (x3-2x)(x2-1) - (x2-1)x = x5 - 4x3 + 3x

As Somos writes, these polynomials can be converted to the Chebyshev polynomials of the second kind by replacing x with 2x everywhere (so for example 3x2 becomes 12x2) and offsetting the index by one (so for example s(5) becomes U(4)).

Congrua of the Form kb2 for Small Integer k

Fermat took up this problem and looked for values of the congruum that are equal to some small integer times a square. He showed that there are no solutions of the form a2±b2, a2±2b2, or a2±3b2; something of the form a2±4b2 is ruled out by the a2±b2 result; this left a2±5b2 as the first open problem.

Somos gives the following four sequences, which jointly are generated by a recurrence, and give solutions that sometimes involve a minus sign (and thus an i in the number which, when squared, is one of the values a2±5b2):

b = W(n) = 0, 1, -12, -2257, 1494696, 8914433905, -178761481355556, -62419747600438859233, ... (A129206)

X(n) = 1, 1, -49, -4799, 4728001, 18618840001, -767067390499249, -54213419267800732799, ... (A129207)

a = Y(n) = 1, 2, 41, 1562, 3344161, -7118599318, 654686219104361, -128615821825334210638, ... (A129208)

Z(n) = 1, 3, 31, 5283, -113279, 21166249443, -518493692732129, 189797666150873887683, ... (A129209)

The first few examples are:

22-5×12 = -1 = i2 ; 22+5×12 = 9 = 32
412-5×122 = 961 = 312 ; 412+5×122 = 2401 = 492
15622-5×22572 = -23030401 = 4799i2 ; 15622+5×22572 = 27910089 = 52832


References

[1] Michael Somos, Step into the Elliptic Realm, 2000 Mar 23.

[2] Michael Somos, In the Elliptic Realm, 2015 Feb 12.


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