Open Contests

a) By taking $x=y=\frac{1}{2}$ we have that $x+y=1$ is an integer and so is $ax+by=\frac{1}{2}(a+b)$, since $a+b$ is even by the assumption.

b) We notice that $ax+by=a(x+y)+(b-a)y$. Assume that $x+y$ and $ax+by$ are integers. Since $a(x+y)$ is an integer, since it is a product of two integers, $(b-a)y$ must be an integer as well. But in the case $b-a=1$ this is not possible, since $y$ is noninteger by the assumption.

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Alternative solution.

a) If $a=b$ then any noninteger numbers $x$ and $y$ whose sum is an integer will be suitable as in this case $ax+ay$ is an integer, as it is a product of two integers $a$ and $x+y$. In the case $a\ne b$ we can take $x=\frac{1}{a-b}$ and $y=\frac{a-b-1}{a-b}$, as in that case $x+y=1$ and $$ax+by=\frac{a}{a-b}+\frac{b(a-b-1)}{a-b}=\frac{(b+1)(a-b)}{a-b}=b+1.$$

b) Assume $x+y=n$ and $ax+by=m$, where $n$ and $m$ are integers. By interpreting this as a system of equations and solving for $x$ and $y$ we obtain $x=\frac{m-bn}{a-b}$ and $y=\frac{an-m}{a-b}$ (as $a$ and $b$ have different parities we have $a-b\ne 0$). If $a-b=1$ then these solutions are integers. Thus we can not guarantee the existence of noninteger numbers with desired properties.