Suppose satisfy show that
From this post on I will try to adopt a more casual tone in expounding the solution, with explanatory text mixing with the solution.
First, we see that the equation does not easily lead itself to non-trivial solution (the term is quite weird), therefore we decide to punish it by repeatedly substituting different values of variables until we can compare the information about the values to find a contradiction.
Setting , and cancelling gives .
Let be any integer, substituting will give us Since the LHS vanishes, we thus have or . Thus we only have to rule out the latter case and the proof is complete.
Notice that the functional equation, though strange in , behaves benignly when only the parity is concerned: in fact, setting , we have a ‘solution’ to it . Thus we want to consider parity. Setting , we have The RHS, incidentally, is always an even number, whereas the LHS is the function value of any even number. Thus maps even numbers to even numbers. (Note that the same cannot be said of odd numbers as is a solution to the equation.)
We aim to utilise the assumption that We let , for convenience. Let in the original equation, we obtain . Now we implant our initial strategy of trashing the function with repeated substitutions.
Let , we find Setting , we thus have thus . We thus know that and are all contained in the image of .
Setting , we have .
Setting , we have . Setting in the place of .
Setting , we have .
Subtracting the two equations above, we get
(This is why we used instead of when , as that would permit us to subtract and reduce one term.)
Upto this point, we haven’t used any property of (we haven’t in fact used the parity condition). But in an Olympiad problem, we should aim to use all the given conditions, as the problems are almost always tailored such that the conditions just suffice for a solution using elementary method. Now we observe that the LHS of the equation (*) is the difference between two terms with distance apart, while the RHS has a factor that is also a difference between two terms with distance apart, with another factor whose absolute value is at least if we can make sure it is not zero. We want it to be . This can be done by inequality.
If , then is a constant on even number (with that value being thus as ) and a constant on odd number (with that value thus being as otherwise would not be in the range of .) Now, we have shown that is also in the image of the function, thus this is a contradiction.
Thus . Consider the set , with its minimal element being attained at . Substitute into (*), and take absolute value.
By the minimality condition, , and because it does not vanish (as the LHS is non-zero)
Multiplying them, we have equality in the face of two inequalities. Thus the two inequalities must have equality holding at the same time, thus . This is impossible since we have shown above that
Comment: The problem is not very difficult, and many alternative methods exist, as the function itself is quite bizarre and repeated substitution per se gives more and more information from which any contradiction would finish the problem off.
The problem is proposed by Qiang Shen, and used in a practice exam at Renhui Jiuzhang MO Tuition Centre (Beijing, China) in 2015.