solution to an equation involving natural numbers

Question

Suppose $a,b$ are real numbers and we have

$$ 1 = b + an \; \; \; \; \forall n \in \mathbb{N} $$

My book says that only solution is $b=0$, $a=1$. But, this does not make sense to me since if we put $n = 1$, we have

$$ 1 = b + a $$

And $a=b=1/2$ is a solution. What is wrong?

Answer 1

For all $n$ implies that we have:

$$1=b+a$$ $$1=b+2a$$

So $a=0$ and $b=1$ is the only solution.

Answer 2

The problem with setting $a=b=\frac12$ is that in this case, the equality $$1=b+an$$

is not true for every $n\in \mathbb N$. In particular, it is not true for $n=2$, since, for $n=2$, you have $b+na=b+2a=\frac12 + 2\cdot \frac12 = \frac32 \neq 1$.

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In fact, for any single value of $n$, you have infinitely many solutions of the equation. But there exists only one solution (the one where $a=0, b=1$) which is a solution for every value of $n$.

This is also fairly easy to show, since:

1. If the equation $1=b+an$ is true for all $n$, then it is true for $n=1$ in particular, meaning that $a+b=1$
2. If the equation $1=b+an$ is true for all $n$, then it is true for $n=2$ in particular, meaning that $2a+b=1$
3. The two equations together ($a+b=1$ and $2a+b=1$) have only one solution, $a=0$ and $b=1$.
4. Therefore, if there is any pair of $a,b$ such that $1=b+an$ for all $n$, then that pair is $(a,b)=(0,1)$.
5. Clearly, $(0,1)$ is a solution for every $n$, since $1=1+0\cdot n$ for every value of $n$.