The number
2
4
is a special number because it is one less than a square (
5
2
) and if it's multiplied by
2
, that answer is also one less than a square (
4
8
+
1
=
4
9
=
7
2
). Find the lowest possible special number greater than
2
4
.
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It was asking for less than 24?
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Yeah my bad It's my first problem . And yes It was supposed to not count 0 and I've edited it
Let the number n satisfies n = a 2 − 1 , 2 n = b 2 − 1 , where a , b are integers. Then
a = 2 b 2 + 1
So b must be odd. Say b = 2 m + 1 . Then a = 2 m 2 + 2 m + 1 = m 2 + ( m + 1 ) 2 .
So m , m + 1 , a must form a Pythagorean Triplet . We know that ( 3 , 4 , 5 ) , ( 2 0 , 2 1 , 2 9 ) , . . . are such triples.
So, next to m = 3 , m = 2 0 ⟹ a = 2 9 , and n = 2 9 2 − 1 = 8 4 0 .
Brute force:
Iterating over x : let y = 2 ( x 2 − 1 ) and test whether ( y − 1 ) is a perfect square.
Yes when x = 1 ( x 2 − 1 = 0 ), x = 5 ( x 2 − 1 = 2 4 ) and x = 2 9 ( x 2 − 1 = 8 4 0 ).
8 4 1 = 2 9 2 and 1 6 8 1 = 4 1 2 .
I assumed that the question was not looking for 0 .
This is an example of a Pell equation , by the way, specifically x 2 − 2 y 2 = − 1
The values 2 4 and 8 4 0 correspond to values of y 2 − 1 in these solutions. The next "special" numbers are 2 8 5 6 0 and 9 7 0 2 2 4 .
All of these solutions are related to approximations to 2 : the values of y x for the four special numbers above are
Special num | x | y | y x |
2 4 | 7 | 5 | 1 . 4 |
8 4 0 | 4 1 | 2 9 | 1 . 4 1 3 7 9 … |
2 8 5 6 0 | 2 3 9 | 1 6 9 | 1 . 4 1 4 2 0 … |
9 7 0 2 2 4 | 1 3 9 3 | 9 8 5 | 1 . 4 1 4 2 1 … |
The linked article gives a lot more information, including how to generate new solutions.
Interesting Pell observation, Chris. Thanks! I notice this sequence has some other interesting properties (and conjectures) in OEIS , by the way.
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let A = x 2 − 1 , then 2 A = 2 ( x 2 − 1 ) = y 2 − 1 , ∴ y 2 = 2 x 2 − 1
So y must be odd and y 2 = 4 k + 1 , then x must be odd too.
Then it is easy to get:
y = 1 , A = 0
y = 7 , A = 2 4
y = 4 1 , A = 8 4 0
y = 2 3 9 , A = 2 8 5 6 0