A simultaneous equation from Putnam

Putnam is one of the most well known undergraduate maths competitions in the world. To some, it is even the hardest maths competition in the world. Although, if you are too scared to try hard problems, you can never improve in maths. With a median score of 0 or 1 out of 120 per year on average, we now attempt this problem. Before looking at the solution, give the problem a go!

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We immediately notice that we can cancel one of the terms on the left hand side by both adding and subtracting. We thus get two new equations that are:

Naturally, we’re going to expand this to see if we get anything nicer. We then yield:

We notice that the top and bottom are very similar… they have the same numbers coefficients just placed in different places. Let us finish multiplying out the left hand side to yield:

Which is valid provided that neither x nor y are equal to 0. It seems that every term is now different but has the same coefficients. At first it may seem like a problem, but then we realise that these are precisely the binomial coefficients. That is, we can add and subtract these from each other once more to get the following:

And therefore:

Which is much simpler! Since 5 is odd and we’re only looking for real solutions, we can now just take the 5th root of each side to yield a pair of linear equations! Thus we now have:

This is now simple to solve. Subtracting and adding equations one last time we get :

as our only real solution. In fact, if we check using online graphing software:

This is indeed our only solution (note that the equations are undefined at 0 thus do not meet and behave asymptotically, meaning they get arbitrarily close to each other without crossing, at the axis)