Arthur’s (in)famous puzzles — David’s solutions

The aim is to find non-negative integer values for A, B and N which satisfy:

A² + B²

A B + σ

= N

where σ = ±1

This can be rearranged as a quadratic equation in A.
      A² + B² = N A B + σ N
      A² – N A B – σ N + B² = 0
So for any solution A_n–1 , B_n–1 there is another solution A_n , B_n in which:
      B_n = B_n–1
      A_n = N B_n–1 – A_n–1
As the problem is invariant under interchange of A and B, we can obtain multiple solutions by swapping A_n and B_n:
      A_n = B_n–1
      B_n = N B_n–1 – A_n–1
For most starting values of A₀ and B₀ this gives an infinite series, but not necessarily so as is evidenced by N = 1, A₀ = 1, B₀ = 0 (This is only a solution for σ = +1.):

n =	0	1	2	3	4	5	6	7	...
A_n =	+1	0	–1	–1	0	+1	+1	0	...
B_n =	0	–1	–1	0	+1	+1	0	–1	...

Let B₀ be the value of B of smallest magnitude for any given value of N for which solutions exist, and A₀ be the corresponding solution for A (for which there will be two solutions). Because the problem is invariant under interchange of A and B, each of the values A₀ is also a valid value of B, and thus cannot be of smaller magnitude than B₀.
      2 A₀ = N B₀ ± √(N ² B₀² – 4 B₀² + 4 σ N)
      | A₀ | ≥ | B₀ |
      \ A₀² ≥ B₀²
The smaller of the two values of A₀ is obtained by choosing the minus sign, thus:
      [ N B₀ – √(N ² B₀² – 4 B₀² + 4 σ N) ]² ≥ 4 B₀²
      N ² B₀² + N ² B₀² – 4 B₀² + 4 σ N – 2 N B₀ √(N ² B₀² – 4 B₀² + 4 σ N) ≥ 4 B₀²
      2 N ² B₀² – 8 B₀² + 4 σ N – 2 N B₀ √(N ² B₀² – 4 B₀² + 4 σ N) ≥ 0
      B₀² (N ² – 4) + 2 σ N ≥ N B₀ √(N ² B₀² – 4 B₀² + 4 σ N)

We restrict consideration situations where both sides will be non–negative (e.g. N > 2, σ = +1), so we can square the inequality:
      B₀⁴ (N ² – 4)² + 4 σ N B₀² (N ² – 4) + 4 N ² ≥ N ² B₀² (N ² B₀² – 4 B₀² + 4 σ N)
      B₀⁴ (N ² – 4)² + 4 σ N B₀² (N ² – 4) + 4 N ² – N ² B₀⁴ (N ² – 4) ≥ 4 σ N³ B₀²
      B₀² (N ² – 4) [B₀² (N ² – 4) + 4 σ N – N ² B₀²] + 4 N ² ≥ 4 σ N³ B₀²
      B₀² (N ² – 4) [ 4 σ N – 4 B₀² ] + 4 N ² – 4 σ N³ B₀² ≥ 0
      B₀² (N ² – 4) (σ N – B₀²) + N ² – σ N³ B₀² ≥ 0
      B₀² [ (N ² – 4) (σ N – B₀²) – σ N³ ] + N ² ≥ 0
      B₀² (4 B₀² – 4 σ N – N ² B₀²) + N ² ≥ 0
      N ² ≥ N ² B₀⁴ + 4 σ N B₀² – 4 B₀⁴
      0 ≥ B₀⁴ (N ² – 4) + 4 σ N B₀² – N ²
This factorises (note σ² = 1) to:
      0 ≥ [B₀²(N + 2) – σ N] [B₀²(N – 2) + σ N]

For σ = +1 and N ≥ 2, the term in the second square bracket is always positive:
      \ B₀²(N + 2) – N ≤ 0

and

B₀²

≤

N + 2

Thus for any value of N ≥ 2 for which there are integer solutions for A and B, there will be a solution in which B = 0 and A² = N.
For N = 1, we still have a solution in which B = 0 and A² = N.
Thus any integer value of A gives an integer value of N which is always a perfect square.

For σ = –1 the discriminant in the quadratic equation solving for A will be negative unless N > 2, so the term in the first square bracket is always positive:
\ B₀²(N – 2) – N ≤ 0

and

B₀²

≤

(N – 2)

1 +

N – 2

This gets smaller for increasing N, starting with the value 3 for N = 3 so that:
      B₀ < √3
i.e. whatever the value of N for which there are solutions there will always be a solution for B₀ = 1.

We now look for those values of N for which there is a solution with B = 1.
We can substitute B = 1 in our quadratic equation for A, and observe that for an integer solution, the discriminant must be a perfect square.
We introduce integers Q and Z such that:
      Z² = N ² – 4 N – 4
      0 = N ² – 4 N – 4 – Z²
      N = 2 ± Q
      Q² = 4 + 4 + Z²
      ( Q – Z ) ( Q + Z ) = 8
             where Q and Z are non–negative integers, and Q > Z. \ Z < 4.
      Z = 0 : Q² = 8
      Z = 1 : Q² = 8 + 1 \ Q = 3 and N = 5
      Z = 2 : Q² = 8 + 4 = 12
      Z = 3 : Q² = 8 + 9 = 17
Therefore the only solutions for B = 1 have N = 5.
Combined with our earlier proof that all sets of solutions for a given N must contain a solution in which B = 1, this proves that the only value of N for which integer solutions for A and B exist is 5.