**Question Corner and Discussion Area**

I have been working on question #4 from this years IMO paper:(Note: the "IMO" is the "International Mathematical Olympiad")4. The positive integers

aandbare such that the numbers 15a+ 16band 16a- 15bare 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
and
where *a*, *b*, *c*, and *d* are positive integers. You
are asked to find what is the smallest that the smaller of and
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 ,
you get . This is guaranteed to
be positive, so the only other thing necessary is that *a* be an
integer, i.e.

must be a multiple of . (Solving forCondition A)

must be a multiple of . (Summarizing: the only conditions on the squares and are that they're positive, satisfy the inequality , and satisfy conditions A and B.Condition B)

Whenever you have a condition like A or B, it's useful to consider prime
factors. The prime factors of are 3 and 7. Let's work with 3 first.
Since 3 divides and 3 also divides , condition A tells
us that 3 must also divide 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.

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