Diophantine Equation

Version 1.1

Introduction

The following diophantine equation was a topic problem from #math. The origin of the problem is unknown to me(landen). To explore ideas and cut down on mistakes I used the software package for number theory, PARI/GP, which is free for noncommercial use. This package is convenient for calculations in algebraic extensions of Q and Z.

2a + 3b = 5c      (1)

The problem can be split into four cases: a is odd with a ≥ 3, a=2, a=1, and a is even with a ≥ 4. Two of the cases can be solved by just trying all possibilities (mod 3) and (mod 8). The remaining cases are a=1, and a is even with a ≥ 4, which take more work.

Proof: Case 1, a is odd with a ≥ 3

Exploring equation (1) (mod 3) we see that a and c must have the same parity in general. So c = 2n+1 Rearranging (1) we have:

2a + 3b = 52n+1 and then (mod 8)
2a = 0 (mod 8); 3b = 1 or 3 (mod 8); 52n+1 = 5 (mod 8)
So (1) fails (mod 8) for this case. This case gives no solutions.
Q.E.D.

Proof: Case 2, a = 2

c must even be since a and c have the same parity in general. However, (mod 8), c must be odd which is a contradiction.
Q.E.D.

Proof: Case 3

2 + 3b = 5c has only one solution, b = 1 and c = 1      (2)

Assume b = 2m+1 and c = 2n+1 with n > 0. They must be odd by the argument in Case 1. Multiplying the equation by 3 and rearranging terms we get:

32(m+1) - 15 52n = -6      (3)

In this form we see that any solutions to (2) must also be among solutions to:

x2 - 15 y2 = -6      (4)

Of course (4) will have many more solutions than (3) but solutions to (4) are easy to study in Z[w], where w is sqrt(15). The only solutions to (4) which have a chance of solving (3) are the ones in which y is a power of 5. One way to show that (3) has no solutions is to show that when y is divisible by 3 it is also divisible by 11. This property of the solutions was first conjectured after using PARI/GP to experiment. It was then easy to prove that conjecture by considering the solutions to (4) (mod 5) and (mod 11). From the theory of binary quadratic forms we know that all the solutions to (4) can be written as:

s(k) = s0uk = x(k) + y(k)w    (5)
k is in Z
w = sqrt(11)
s0 = 3-w a solution of (4) and u = 4+w, the fundamental unit of Z[w].
norm(x(k) + y(k)w) = x(k)2 - 15 y(k)2 = -6
PARI/GP can compute solutions using (5) directly. Here are solutions for a few values of k:

k = -12;  s(k) = (194888982963 - 50320119025*w);  norm(s(k)) = -6
k = -11;  s(k) = (24754146477 - 6391493137*w);  norm(s(k)) = -6
k = -10;  s(k) = (3144188853 - 811826071*w);  norm(s(k)) = -6
k = -9;  s(k) = (399364347 - 103115431*w);  norm(s(k)) = -6
k = -8;  s(k) = (50725923 - 13097377*w);  norm(s(k)) = -6
k = -7;  s(k) = (6443037 - 1663585*w);  norm(s(k)) = -6
k = -6;  s(k) = (818373 - 211303*w);  norm(s(k)) = -6
k = -5;  s(k) = (103947 - 26839*w);  norm(s(k)) = -6
k = -4;  s(k) = (13203 - 3409*w);  norm(s(k)) = -6
k = -3;  s(k) = (1677 - 433*w);  norm(s(k)) = -6
k = -2;  s(k) = (213 - 55*w);  norm(s(k)) = -6
k = -1;  s(k) = (27 - 7*w);  norm(s(k)) = -6
k = 0;  s(k) = (3 - w);  norm(s(k)) = -6
k = 1;  s(k) = (-3 - w);  norm(s(k)) = -6
k = 2;  s(k) = (-27 - 7*w);  norm(s(k)) = -6
k = 3;  s(k) = (-213 - 55*w);  norm(s(k)) = -6
k = 4;  s(k) = (-1677 - 433*w);  norm(s(k)) = -6
k = 5;  s(k) = (-13203 - 3409*w);  norm(s(k)) = -6
k = 6;  s(k) = (-103947 - 26839*w);  norm(s(k)) = -6
k = 7;  s(k) = (-818373 - 211303*w);  norm(s(k)) = -6
k = 8;  s(k) = (-6443037 - 1663585*w);  norm(s(k)) = -6
k = 9;  s(k) = (-50725923 - 13097377*w);  norm(s(k)) = -6
k = 10;  s(k) = (-399364347 - 103115431*w);  norm(s(k)) = -6
k = 11;  s(k) = (-3144188853 - 811826071*w);  norm(s(k)) = -6
k = 12;  s(k) = (-24754146477 - 6391493137*w);  norm(s(k)) = -6
k = 13;  s(k) = (-194888982963 - 50320119025*w);  norm(s(k)) = -6

By factoring y(k) when it is a multiple of 5, it was conjected that y(k) is also divisible by 11 and thus is not a power of 5. When s(k) is computed (mod 5) and (mod 11) the results are periodic with a period of ten. Notice that

u5 = 15124 + 3905*w
u5 = 1 (mod 5) Proves solutions repeat with a period of 5 (mod 5).
u5 = -1 (mod 11) Proves solutions repeat with a period of 10 (mod 11).
But every 5th term (mod 11), the coefficient of w (mod 11) in u5 = 0, so is divisible by 11 also.

y(3) is 55 which is divisible by 5 and 11. This means that divisiblity by 5 and 11 occur together at y(3 + k*5). Thus y is never a power of 5.
Q.E.D.

As a check for errors, PARI/GP calculated the solutions (mod 5) and (mod 11). Here are some results:

** means coefficient of w, y(k) is divisible by 5 and 11 both

k = -7; s(k) = 2 + 0*w (mod 5); s(k) = 7 + 0*w  (mod 11) **
k = -6; s(k) = 3 + 2*w (mod 5); s(k) = 6 + 7*w  (mod 11)
k = -5; s(k) = 2 + w   (mod 5); s(k) = 8 + w    (mod 11)
k = -4; s(k) = 3 + w   (mod 5); s(k) = 3 + w    (mod 11)
k = -3; s(k) = 2 + 2*w (mod 5); s(k) = 5 + 7*w  (mod 11)
k = -2; s(k) = 3 + 0*w (mod 5); s(k) = 4 + 0*w  (mod 11) **
k = -1; s(k) = 2 + 3*w (mod 5); s(k) = 5 + 4*w  (mod 11)
k =  0; s(k) = 3 + 4*w (mod 5); s(k) = 3 + 10*w (mod 11)
k =  1; s(k) = 2 + 4*w (mod 5); s(k) = 8 + 10*w (mod 11)
k =  2; s(k) = 3 + 3*w (mod 5); s(k) = 6 + 4*w  (mod 11)
k =  3; s(k) = 2 + 0*w (mod 5); s(k) = 7 + 0*w  (mod 11) **
k =  4; s(k) = 3 + 2*w (mod 5); s(k) = 6 + 7*w  (mod 11)
k =  5; s(k) = 2 + w   (mod 5); s(k) = 8 + w    (mod 11)
k =  6; s(k) = 3 + w   (mod 5); s(k) = 3 + w    (mod 11)
k =  7; s(k) = 2 + 2*w (mod 5); s(k) = 5 + 7*w  (mod 11)
k =  8; s(k) = 3 + 0*w (mod 5); s(k) = 4 + 0*w  (mod 11) **

Proof: Case 4

a is even with a ≥ 4 has only the solution a=4,b=2,c=2

Assume a is even with a ≥ 4. Then b = 2m and c = 2n by considering equation (1) (mod 8). Rearranging (1) we have:

 2a = 52n - 32m = (5n - 3m)(5n + 3m)

By looking at the sum and difference of the two factors on the right, we see that they have 2 as their greatest common divisor. This means that

(5n - 3m) = 2, and therefore n=1 and m=1 by Case 3, so (a,b,c) = 4,2,2)

Q.E.D.

http://efnet-math.org/math_tech/dioph2.htm