Find two integers so that X^2 + 2XY + 33Y^2 = r where r = 12345678987654321

I already have X = 111 111 111 and Y = 0 But i'd like to find at least another one.

Thank you very much for the help :)

Hello,

With r=12,345,678,987,654,321 et s=111,111,111
We have s²=r.

x² + 2xy + 33y² = r
x² + 2xy + y² + 32y² = r
(x + y)² + 32y² = r

If y=0, then
   (x + 0)² + 32×0² = r
   x² = s²
   x = ±s

So you get two trivial solutions by setting y=0.
   (x; y) = (-111,111,111 ; 0)
   (x; y) = (111,111,111 ; 0)

Now:
(x + y)² + 32y² = r
(x + y)² – (r – 32y²) = 0

Supposing r–32y²≥0 :
(x + y)² – (r – 32y²) = 0
(x + y)² – [√(r – 32y²)]² = 0
[x + y – √(r – 32y²)][x + y + √(r – 32y²)] = 0

Thus through null factor law:
x + y = ±√(r – 32y²)

Since x and y are integers,
√(r–32y²) is also an integer.
r–32y² is a perfect square.

Thus, let that perfect square be a²:
r – 32y² = a²
s²/y² – 32 = a²/y²

s = 111,111,111 = 3² × 37 × 12345679

→→→ Here we will make an assumption to try to solve the problem

We will assume y perfectly divides s. y is then a divisor of s and we would have the possibilities:
y ∈ {1; 3; 9; 37; 111, 333; 12345679; 37037037; 111111111}

Then
s/y ∈ {111111111; 37037037; 12345679; 333; 111; 37; 9; 3; 1}

Then (s/y)²–32 ∈ {
  12345678987654289;
  1371742109739337;
  152415789971009;
  110857;
  12289;
  1337;
  49;
  -23;
  -31}

Among these values, only 49 is a perfect square.
It is reached when y=12345679

Thus, if y=12345679
r – 32y² = a² = 7²×12345679² = 86419753²

And since
x + y = ±√(r – 32y²)
x + y = ±86419753
x = -y ± 86419753

First possibility:
x = -12345679 – 86419753 = -98765432

And:
x² = (-98765432)² = 9754610558146624
2xy = -2438652639536656
33y² = 5029721069044353
sum = 12345678987654321
QED.

Second possibility:
x = -12345679 + 86419753 = 74074074