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: Contradiction and conclusion

Previous step Next step

d\ \text{even contradicts}\ d=2k^2-2k+1\ \text{odd}

Detailed analysis

This contradicts Step 2, where $d=2k^2-2k+1$ was shown to be odd for every $k$ . Hence $2d-1,5d-1,13d-1$ cannot all be perfect squares, so some pair $a,b\in\{2,5,13,d\}$ satisfies that $ab-1$ is not a perfect square.