Find all possible pairs (x, y) of prime numbers, with x less than or equal to y, and each of x and y being less than 2023, such that:
• y divides x2 + 8 and,
• x divides y2 + 8
Provide valid reasoning as to why there are no further solutions.
*** Extra credit for an analytical solution.
Note: This puzzle is adapted from a problem appearing in the Zhautykov Olympiad.
This is an ingenious problem.
The first point to understand is that the small values need to be ignored in finding a general approach to the larger ones.
Let (x^2+8)=ay, (y^2+8)=bx
By inspection we can get the solution {x,y} = {2,2}, as well as the only case where a=b over the primes.
Since we are solving for odd primes, let x=(2c-1) and y=(2d-1) such that:
c=1/2(sqrt(ay-8)+1), d=1/2(sqrt(bx-8)+1), as we only want the positive roots.
Try small values n=1/2(sqrt(bx-8)+1)
1: c=1,3,9
2: c=17/b
3: c=33/b
4: c=57/b
5: c=89/b
etc.
Note that al of these after 1 and 3 are of the form (2n-1)^2+8. It then becomes clear that there is a four-way symmetry between a,b,x, and y. Call these values the 'seed values'.
Concentrating on x, though, industry provides a series of quartics:
x^4+16x^2-4xz^2+4xz-9x+72 = 0
x^4+16x^2-36xz^2+36xz-81x+136 = 0
x^4+16x^2-324xz^2+324xz-729x+712 = 0
and a complicated multinomial for larger values, so let's just use the seed values:
x^4+16x^2+x(-1156z^2+1156z-2601)+2376 = 0
x^4+16x^2+x(-4356z^2+4356z-9801)+8776 = 0
x^4+16x^2+x(-12996z^2+12996z-29241)+26056 = 0
x^4+16x^2+x(-31684z^2+31684z-71289)+63432 = 0
etc.
Given that x appears in every part of each expression but the constant, x must divide the constant term:
72 = 2^3×3^2
136 = 2^3×17
712 = 2^3×89
2376 = 2^3×(3^3×11) (297)
8776 = 2^3×1097
26056= 2^3×3257
63432= 2^3×3^2×881
so x can only be selected from the odd prime factors that make up these numbers.
Testing these, we get the valid solutions {x,y,a,b} = {3,17,1,99}, {89, 881, 9, 8721}, {881, 8721, 89, 86329}.
Given the values {1, 9, 89, 881, 8721, 86329,...}, the recurrence a(n) =a(n) = ceiling(((3-sqrt(6))*(5+2*sqrt(6))^n)/6), is easily identified, and a complete solution set for all possible {x,y,a,b} be written as x=a(n), y=a(n+1), a=a(n-1), b=a(n+2).
So the specific answer to the problem is: {x,y} = {2,2}, {3,17}, {89, 881}.
Edited on March 19, 2023, 3:40 am
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Posted by broll
on 2023-03-19 03:38:45 |
Possible Solution
