1.5 KiB
1.5 KiB
| id | title | challengeType | forumTopicId | dashedName |
|---|---|---|---|---|
| 5900f3ae1000cf542c50fec1 | Problem 66: Diophantine equation | 5 | 302178 | problem-66-diophantine-equation |
--description--
Consider quadratic Diophantine equations of the form:
x2 – Dy2 = 1
For example, when D=13, the minimal solution in x is 6492 – 13×1802 = 1.
It can be assumed that there are no solutions in positive integers when D is square.
By finding minimal solutions in x for D = {2, 3, 5, 6, 7}, we obtain the following:
32 – 2×22 = 1
22 – 3×12 = 1
92 – 5×42 = 1
52 – 6×22 = 1
82 – 7×32 = 1
22 – 3×12 = 1
92 – 5×42 = 1
52 – 6×22 = 1
82 – 7×32 = 1
Hence, by considering minimal solutions in x for D ≤ 7, the largest x is obtained when D=5.
Find the value of D ≤ 1000 in minimal solutions of x for which the largest value of x is obtained.
--hints--
diophantineEquation() should return a number.
assert(typeof diophantineEquation() === 'number');
diophantineEquation() should return 661.
assert.strictEqual(diophantineEquation(), 661);
--seed--
--seed-contents--
function diophantineEquation() {
return true;
}
diophantineEquation();
--solutions--
// solution required