Week 41

Example 1. Find all pairs of positive integers x and y that satisfy 5x − 7y = 1 and (x/y) > √2.

Example 2. Suppose that every point in the plane is coloured either red or blue. Prove that there exists an isosceles right triangle all of whose vertices have the same colour.

Solution 1. An obvious solution to 5x − 7y = 1 is (x, y) = (3, 2).

Therefore, for any integer solution (x, y) to this equation, we have

5x − 7y = 1 = 5 × 3 – 7 × 2

or, 5(x − 3) = 7(y − 2).

Since 5 and 7 are distinct primes, the last relation implies that x − 3 = 7t and y − 2 = 5t for some integer t.

This mean that x = 7t + 3 and y = 5t + 2.

In addition, we are given that (x/y) > √2.

Squaring both sides and then substituting, we get

x² > 2y²

or, (7t + 3)² > 2(5t + 2)²

or, 2(5t + 2)²  − (7t + 3)² < 0

or, 50t² + 40t + 8 − 49t² − 42t – 9 < 0

or, t² − 2t – 1 < 0

or, (t − 1)² – 2 < 0

or, (t − 1)² < 2.

Only possible solutions are t = 0, 1 or 2.

Therefore we have only three possible pairs, namely:

(x, y) = (3, 2), (10, 7) and (17, 12).

Solution 2. Let A and B be two points of the same colour, say red.

Draw the square ABCD as indicated and let O be its centre (the meeting point of the diagonals).

If point C is red, then ∆ABC is an isosceles right-angled triangle with all vertices red, and we are done.

We can thus assume that C is coloured blue, and similarly, we can assume that D is blue (if D is coloured red, then ∆ABD is an isosceles right-angled triangle with all vertices red).

Finally, if O is red, then ∆AOB is an isosceles right-angled triangle with all vertices red, and if O is blue, then ∆COD is an isosceles right triangle with all vertices blue.

This completes the proof.