The equation a
^{2} + b
^{2} + c
^{2} = d
^{2} has an infinite number of solutions in positive integers, however, if we restrict a and b to ≤ 3600 then there are only a finite set of solutions.
Find the values a ≤ b ≤ 3600 which give the largest number of solutions.
Since nobody else has attempted this:
Let a^2+b^2=d^2c^2.
This is a convenient place to start since LHS presents as a sum of two squares, for which there are known methods of maximising representations, and known limitations.
If a=b=3600^2, then a^2+b^2=25920000.
Maximal representations near this number are easily established, and 5928325 = 5^2×13×17×29×37 stands out as a local maximum below 25920000. Essentially, numbers of this type like lots of different prime factors of the form (4k+1) and somewhat like powers of those factors. Powers of the form 2^n do not help. Powers of primes of the form (4k+3) do not help if even, and are forbidden, if odd.
e.g: a^2+b^2 = 5928325 has 48 positive representations, while a^2+b^2 = 160225 (omitting 37) has 24, and a^2+b^2 = 1185665 (omitting a 5) has 32.
RHS obeys a different rule. d^2c^2 has the most representations whenever it has the most divisors, provided that d^2c^2 cannot equal 4k2.
e.g: d^2c^2=10810800=2^4×3^3×5^2×7×11×13 has 144 representations, while d^2c^2=5405400 has 96 (omitting a 2), and d^2c^2=1544400
(omitting 7) has 72.
So the problem is equivalent to finding a compromise between most divisors for the difference of squares and most factors of form (4k+1) for the sum of squares.
How many representations can the difference of squares have?
21621600 has 192 representations, but also many factors of form (4k+3) to odd powers and powers of 2. However, we can treat 192 as a ceiling for the difference of squares.
In summary:
1. a^2+b^2 cannot exceed 48 representations.
2. d^2c^2 cannot exceed 192 representations, for a total of 9216.
3. Let x be the desired number. Then:
(a) every factor of x is in {5,9,13,17,25,29,37} and powers of 2. However, 9 and powers of 2 do not improve the number of representations for the sum of squares.
(b) x cannot be of the form 4k+2.
(c) x should minimise use of higher powers of its factors as far as possible.
We could start with 5928325 itself. This has 24 representations (or 48 including {a,b,} switched). d^2c^2=5928325 has 24 representations.
24*24 = 576, the number I gave earlier.
If we include {a,b,} switched then the total becomes 1152.
However, while writing the above, I discovered 10670985 = 3^2×5×13×17×29×37. This has 32 representations (so a loss of 16, counting switches) for a^2+b^2. But it has 48 representations as a difference of squares.
32*48=1536
Still plenty of room for improvement!
Edited on May 30, 2024, 8:09 am

Posted by broll
on 20240530 07:47:28 