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I accidentally deleted this post, and fixed the example and reproduced it.

In number theory, it is well known that the sum of two unit fractions with prime denominators is:

1/p + 1/q = (p + q)/(pq)

I propose the following conjecture:

Conjecture

Let p and q be distinct prime numbers.
Define:

1/p + 1/q = a/b

Then the reversed sum:

1/a + 1/b ≠ 1/r for any integer r.

In other words, the sum of the reciprocals of the resulting numerator and denominator never results in a unit fraction again.

Examples:

p q a = p+q b = p×q 1/a + 1/b Equals 1/r?
2 3 5 6 11/30 No
2 5 7 10 17/70 No
2 7 9 14 23/126 No
2 11 13 22 35/286 No
2 13 15 26 41/390 No
2 17 19 34 53/646 No
2 19 21 38 59/798 No
2 23 25 46 71/1150 No
2 29 31 58 89/1798 No
2 31 33 62 95/2046 No
2 37 39 74 113/2886 No
2 41 43 82 125/3526 No
2 43 45 86 131/3870 No
2 47 49 94 143/4606 No
2 53 55 106 161/5830 No
2 59 61 118 179/7198 No
2 61 63 122 185/7686 No
2 67 69 134 203/9246 No
2 71 73 142 215/10366 No
2 73 75 146 221/10950 No

Reformulation

Assume:

1/p + 1/q = a/b = (p + q)/(pq)
1/a + 1/b = (a + b)/(ab)

Suppose this equals 1/r, then:

ab = r(a + b)

But substituting a = p + q and b = pq, it results in a contradiction in simplification to a clean 1/r.

Conclusion

This pattern appears consistent under all tested values. While seemingly simple, it offers an interesting glimpse into the behavior of prime-based unit fractions.

I invite any counterexamples, comments, or proofs.

Proposed by Naufal R.P.

Bill Dubuque
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Naufal Rakha Putra
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  • Since a,b are supposed to be primes, the two factors on the right must be the same pair, so either $a =a+b$ or $b=a+b$ – Roland F Jun 21 '25 at 05:45
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    @RolandF If I understand the OP correctly, $a$ and $b$ are not supposed to be primes but, rather, the numerator and denominator of the sum of the reciprocal of the distinct primes $p$ and $q$, i.e., $a=p+q$ and $b=pq$, as indicated in their statement of "1/p + 1/q = a/b" and in their table header. As for the OP's conjecture, it's true and relatively easy to prove. – John Omielan Jun 21 '25 at 05:50
  • Duplicate $\ (p,q)=1\Rightarrow (pq+p+q,,pq(p+q)) = 1,$ follows easily from many prior posts. More generally $,(a,b)=1\Rightarrow (a+b,ab) = 1\ \ $ – Bill Dubuque Jun 21 '25 at 06:54

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