Find five positive integers whose reciprocals sum to 1

One can first try to compose the sum of 1 with smaller number of reciprocals. Let&#x27;s start with smallest positive integers. 1&#x2F;2 + 1&#x2F;3 &lt; 1, but 1&#x2F;2 + 1&#x2F;3 + 1&#x2F;4 &gt; 1, and the part 1 - (1&#x2F;2 + 1&#x2F;3) is actually 1&#x2F;6. So, with 3 positive integers we can have 1&#x2F;2 + 1&#x2F;3 + 1&#x2F;6 = 1.<p>Now we can try to replace 1&#x2F;6 with smaller reciprocals, and we need exactly 3 of them, so the problem sort of repeats itself - now we need to find 3 such integers, that their reciprocals sum to 1&#x2F;6. This is the same problem as we just solved, only scaled 1&#x2F;6 times, so instead of 1&#x2F;2, 1&#x2F;3 and 1&#x2F;6 we can have them divided by 6 - so, 1&#x2F;12, 1&#x2F;18 and 1&#x2F;36.<p>So now we have five numbers - 1&#x2F;2, 1&#x2F;3, 1&#x2F;12, 1&#x2F;18 and 1&#x2F;36 - summing to 1 and their reciprocals are integers.<p>You suspect that&#x27;s not the only solution - after all, original 1&#x2F;2 + 1&#x2F;3... was selected rather arbitrarily, as the largest numbers with reciprocals still less than 1.

I came at a solution doing the following:<p>Well, I know this:<p><pre><code> 1 = 1&#x2F;2 + 1&#x2F;2 </code></pre> We&#x27;re trying to get to 5 terms. So let&#x27;s split 1&#x2F;2 into two terms.<p><pre><code> 1 = 1&#x2F;2 + 1&#x2F;4 + 1&#x2F;4 </code></pre> Hmm, let&#x27;s do that again - this time splitting 1&#x2F;4.<p><pre><code> 1 = 1&#x2F;2 + 1&#x2F;4 + 1&#x2F;8 + 1&#x2F;8 </code></pre> Ok, I&#x27;m pretty close. Can I split 1&#x2F;8 into 1&#x2F;9 and some mystery term?<p><pre><code> 1 = 1&#x2F;2 + 1&#x2F;4 + 1&#x2F;8 + 1&#x2F;9 + 1&#x2F;N </code></pre> Well, that&#x27;s not hard to solve. N = 72.<p>1 = 1&#x2F;2 + 1&#x2F;4 + 1&#x2F;8 + 1&#x2F;9 + 1&#x2F;72