I have been working on question #4 from this years IMO paper:(Note: the "IMO" is the "International Mathematical Olympiad")
4. The positive integers a and b are such that the numbers 15a + 16b and 16a - 15b are both squares of positive integers. Find the least possible value that can be taken by the minimum of these two squares.
Can anyone help me solve this?
The conditions given can be translated into 15a + 16b = c^2 and 16a - 15b = d^2 where a, b, c, and d are positive integers. You are asked to find what is the smallest that the smaller of c^2 and d^2 can be.
Let's try and translate all of this information into conditions on c and d. We know they have to be positive integers. The remaining conditions are that a and b have to be positive integers. So, let's solve the above system of equations to express a and b in terms of c and d.
If you do this, and use the fact that 15^2 + 16^2 = 21^2, you get 21^2 a = 15 c^2 + 16 d^2. This is guaranteed to be positive, so the only other thing necessary is that a be an integer, i.e.
15 c^2 + 16 d^2 must be a multiple of 21^2. (Condition A)Solving for b gives 21^2 b = 16 c^2 - 15 d^2. This means that c^2 must be greater than (15/16)d^2, and that
16 c^2 - 15 d^2 must be a multiple of 21^2. (Condition B)Summarizing: the only conditions on the squares c^2 and d^2 are that they're positive, satisfy the inequality c^2 > (15/16) d^2, and satisfy conditions A and B.
Whenever you have a condition like A or B, it's useful to consider prime factors. The prime factors of 21^2 are 3 and 7. Let's work with 3 first. Since 3 divides 21^2 and 3 also divides 15 c^2, condition A tells us that 3 must also divide 16 d^2 and hence must divide d.
Now I'm going to stop giving you the solution and see if you can work out the rest for yourself. You should be able to show that Conditions A and B together are equivalent to a very nice and simple condition on c and d, so that you can tell exactly which numbers are possible values for c and d, and hence from that determine what the smallest possible value is for one of them.
Post another message here if you get stuck further along, or if something I said here wasn't clear.