An Easy IMO Problem!

We consider the combinations $2d$ , $5d$ , and $13d$ for $a$ and $b$ .

Now $2d-1$ , $5d-1$ , and $13d-1$ should be perfect squares.

All perfect squares are $\equiv 0,1,4,9 \pmod{16}$ .

So $d \equiv 1, 5, 9, 13\pmod{16}$

Case 1: $d \equiv 1 \pmod{16}$

$2d-1 \equiv 1 \pmod{16}$

$2d-1$ is a perfect square.

$5d-1 \equiv 4 \pmod{16}$

$5d-1$ is a perfect square.

$13d-1 \equiv 12 \pmod {16}$

$13d-1$ is not a perfect square.

Case 2: $d \equiv 5 \pmod{16}$

$2d-1 \equiv 9 \pmod{16}$

$2d-1$ is a perfect square.

$5d-1 \equiv 8 \pmod{16}$

$5d-1$ is not a perfect square.

Case 3: $d \equiv 9 \pmod{16}$

$2d-1 \equiv 1 \pmod{16}$

$2d-1$ is a perfect square.

$5d-1 \equiv 12 \pmod{16}$

$5d-1$ is not a perfect square.

Case 4: $d \equiv 13 \pmod{16}$

$2d-1 \equiv 9 \pmod{16}$

$2d-1$ is a perfect square.

$5d-1 \equiv 0 \pmod{16}$

$5d-1$ is a perfect square.

$13d-1 \equiv 8 \pmod{16}$

$13d-1$ is not a perfect square.

Therefore, there are $\boxed{0}$ possible values for $d$ .

Edit: In some parts I may have said "Is a perfect square." Not necessarily true but what this means is "Every perfect square can be either $0, 1, 4, 9 \pmod{16}$ . But its converse is not necessarily true."

let 2d - 1 = a, 5d - 1 = b, 13d - 1 = c where a,b,c are perfect squares solve for d and we'll get the following equations,

5a -2b = -3

13b-5c=-8

13a - 2c = -11

three simultaneous linear equations, convert the matrix in row-echelon form and it becomes quite clear that there are no solutions..

see (mod.16) the numbers. the quadratic residues (mod.16) are 0,1,4,9. suppose that $2d-1,5d-1,13d-1$ are perfect squares. for $2d-1\equiv 0,1,4,9 (mod.16)$ , we have $d \equiv 1,5,8 or 13 (mod.16)$ . if d is 1 or 13, then 13d - 1 is not one of these values. If d is 5 or 9, then 5d - 1 is not one of these values.