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 5 of 6: Case

d\equiv1,13\pmod{16}

Previous step Next step

13d-1\pmod{16}\notin\{0,1,4,9\}

Detailed analysis

Direct computation gives $13d-1\equiv12\pmod{16}$ when $d\equiv1\pmod{16}$ , and $13d-1\equiv8\pmod{16}$ when $d\equiv13\pmod{16}$ ; neither $8$ nor $12$ lies in $\{0,1,4,9\}$ . So $13d-1$ is not a perfect square, and $(a,b)=(13,d)$ works.