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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.

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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 $7$.

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  • $\begingroup$ I think the $a_i$ need not be powers of the same prime in the formulation of the question. $\endgroup$ – user757601 Mar 26 '20 at 6:14
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    $\begingroup$ They need not be. How does that affect the answer? $\endgroup$ – Haran Mar 26 '20 at 6:22
  • $\begingroup$ 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 $p$. $\endgroup$ – miracle173 Mar 26 '20 at 6:58
  • $\begingroup$ Yep, that is the same as my answer. $\endgroup$ – Haran Mar 26 '20 at 7:04
  • $\begingroup$ @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 :) $\endgroup$ – Peter Mar 26 '20 at 10:55

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