Problem 1

Let

d

be any positive integer not equal to

2

5

, or

13

. Show that one can find distinct elements

a

and

b

in the set

\{2, 5, 13, d\}

such that

ab - 1

is not a perfect square.

Step 1 of 6: Reformulate the claim

Previous step Next step

In plain words

Perfect squares occupy only a few residue classes modulo any fixed number, so checking $d$ modulo $4$ and then modulo $16$ should already force a contradiction in every case.

2d-1,\ 5d-1,\ 13d-1

Detailed analysis

It suffices to show that for every $d\notin\{2,5,13\}$ , at least one of the three numbers $2d-1$ , $5d-1$ , $13d-1$ (obtained by pairing $d$ with each of $2,5,13$ ) fails to be a perfect square; that pair is then the required $a,b$ .