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 2 of 6: Squares are

0

1

modulo

4

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13d-1\equiv d-1\pmod4

Detailed analysis

Since $13\equiv1\pmod4$ , we get $13d-1\equiv d-1\pmod4$ . A perfect square is $0$ or $1$ modulo $4$ , so $13d-1$ can only be a square when $d\equiv1$ or $2\pmod4$ . Hence if $d\equiv0$ or $3\pmod4$ , $13d-1$ is not a perfect square, and we may take $(a,b)=(13,d)$ .