# Egyptian fractions with prime power denominators summing to 1?

Inspired by

On the decomposition of $1$ as the sum of Egyptian fractions with odd denominators - Part II

Can we solve the equation $$1=\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\cdots +\frac{1}{a_n}$$ where $$a_1,a_2,a_3,\cdots ,a_n$$ are distinct prime powers (primes are allowed) ?

I tried a modified greedy-algorithm with different starting vectors and brute force with $$6$$ entris with limit $$200$$, but did not find a representation.

No, it is impossible. Let $$p \mid a_1$$. Since all terms are distinct, we have some term to be the maximum power of $$p$$ (could be $$a_1$$). Let it be $$a_i=p^k$$.

Now, when we take LCM of denominators to add the fractions, the common denominator is clearly divisible by $$a_i=p^k$$. For each fraction except the one with denominator $$a_i$$, when we take it to common denominator, the new numerator will be divisible by $$p$$. However, the fraction $$\frac{1}{a_i}$$ alone will not have the new numerator divisible by $$p$$ since $$a_i$$ is the highest power of $$p$$ dividing the common denominator.

Now, the numerator is sum of terms divisible by $$p$$ plus one term not divisible by $$p$$. This shows that it is not divisible by $$p$$. However, the denominator is divisible by $$p$$. Thus, it is impossible for the sum to be $$1$$.

The only solution would be $$(a_1,n)=(1,1)$$ where there are no such primes.

P.S. An example might clarify my answer. Take the following: $$\frac{1}{3}+\frac{1}{7}+\frac{1}{7^2}=\frac{7^2}{3 \cdot 7^2}+\frac{3 \cdot 7}{3 \cdot 7^2}+\frac{3}{3 \cdot 7^2}$$

Notice since $$7^2$$ is the highest power of $$7$$, all fractions except $$\frac{1}{7^2}$$ has new numerator divisible by $$7$$, thus when you add the numerators, the sum isn't divisible by $$7$$.

• I think the $a_i$ need not be powers of the same prime in the formulation of the question. – user757601 Mar 26 '20 at 6:14
• They need not be. How does that affect the answer? – Haran Mar 26 '20 at 6:22
• You are 100% right, but maybe a nuance simpler stated: From multiplying $\sum\limits_{i=1}^n\frac { 1 } {a_i} =1$ by $\text{lcm}(a_1,\ldots,a_n)$ follows $\sum\limits_{i=1}^n\frac { \text{lcm}(a_1,\ldots,a_n) } {a_i} =\text{lcm}(a_1,\ldots,a_n)$. All terms but $\frac { \text{lcm}(a_1,\ldots,a_n) } {a_1}$ are divisible by $p$. – miracle173 Mar 26 '20 at 6:58
• Yep, that is the same as my answer. – Haran Mar 26 '20 at 7:04
• @Haran I knew that you would find it out. But I didn't expect such a simple proof. Well done ! (+1 and accept). I thought about this approach but became victim of the fallacy that the sum can be divisble by all the primes, I did not think it to the end :) – Peter Mar 26 '20 at 10:55