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 6 of 6: Case

d\equiv5,9\pmod{16}

and conclusion

Previous step Next step

5d-1\pmod{16}\notin\{0,1,4,9\}

Detailed analysis

Similarly, $5d-1\equiv8\pmod{16}$ when $d\equiv5\pmod{16}$ and $5d-1\equiv12\pmod{16}$ when $d\equiv9\pmod{16}$ ; again neither value is a quadratic residue mod $16$ . So $5d-1$ is not a perfect square and $(a,b)=(5,d)$ works. Every residue class of $d$ has now been handled, so the claim holds for all $d\notin\{2,5,13\}$ .