Minimum number of fractions to be summed up to $\frac45$

Question

What is the minimum number of fractions having numerator 1 and a natural number as denominator to be summed up to $\frac 45$?
I have tested with 2 fractions: $\frac1a + \frac1b = \frac45$ and get into the diophantine equation: $5(a+b)=4ab$ and it seems this should have some solutions but can't find one!!

Answer 1

Considering prime factorization gives that $5 \mid ab$ and, by relabeling if necessary, we can assume that $5 \mid a$, and in particular $\frac{1}{a} \leq \frac{1}{5}$. Substituting in the two-fraction equation gives $b \leq \frac{5}{3}$, so we must have $b = 1$, but this gives $\frac{1}{a} + 1 = \frac{4}{5}$, and the solution $a$ to this equation is negative, so there is no decomposition of $\frac{4}{5}$ as a sum of two such fractions.

On the other hand, if we permit ourselves three such fractions, we can see immediately that one of the fractions must be $\frac{1}{2}$ or $\frac{1}{3}$. (If not, each of the three fractions would be $\leq \frac{1}{4}$ and so would have sum $\leq 3 \cdot \frac{1}{4} < \frac{4}{5}$.) So, this reduces the problem to finding decompositions of either $\frac{3}{10}$ or $\frac{7}{15}$ into two such fractions. There turn out to be two solutions. (Incidentally, fractions of the form $\frac{1}{n}$, where $n$ is a positive integer, are often called Egyptian fractions.)

Answer 2

The equation $4/5 = 1/a + 1/b$ doesn't have any solutions where $a$ and $b$ are positive integers. To prove this, note that either $1/a$ or $1/b$ must be between $2/5$ and $4/5$, i. e. one of $a$ or $b$ must be between $5/4$ and $5/2$. But the only integer in this range is 2, and if $a = 2$ then the corresponding $b$ isn't an integer.

So to express $4/5$ as a sum of unit fractions you need at least three fractions. Can you find a way to do it with exactly three?

Answer 3

Adding to the existing answers, here is another reason why 2 fractions is not enough. With some algebra, you can rearrange your Diophantine equation to:

$$a = \dfrac{5b}{4b -5}$$

Now, varying $b$, starting at $b = 1$ we get the following values for $a$, rounded to two decimals: $-5, 3.33, 2.14, 1.82, \dots$

After the last value, which is less than $2$, clearly we can never get an integer, because the function is monotonically decreasing as $b$ increases, but it is always greater than $\frac54$, which is itself greater than $1$.

Answer 4

Let $a+b=4k$, $ab=5k$, where $k$ is a positive integer.

Then, $b=4k-a$, and substitute into $ab=5k$, we get:

$a(4k-a)=5k$

$a^2-4ka+5k=0$

Then let $ \Delta = 16k^2-20k=(4k)^2-20k=m^2$, where $m$ is another positive integer

And we have $\frac{(8k-5)^2-25}{4}=m^2$

Clearly $(8k-5)^2-25$ should be a perfect square number.

When $k=1$, $9-25<0$, impossible

When $k=2$, $121-25=96$ is not perfect square

When $k\ge3$, $(8k-5)^2-25>(8k-6)^2$.

So there's indeed no positive integer solution to $5(a+b)=4ab$.