Solve the following word/letter/number equations:

For there are three that bear record in heaven, the Father, the Word, and the Holy Ghost: and these three are one.

And there are three that bear witness in earth, the Spirit, and the water, and the blood: and these three agree in one.

A certain amount of creativity is expected. There is more than one solution.

The most satisfying answers, which is what I am looking for, are unlikely to be found by a computational only approach and are not meant to be hard. The hints may be harder to follow the original problem as they are a bit cryptic and direct towards a particular set of answers.

Hint 1:

The answer should concur with the sense of the words.

Hint 2:

Letter position counts

Hint 3:

...or not!

Hint 4 (strengthened):

How much is one worth in total?

Hint 5 (strengthened):

"and", what does that mean? "and", does it always mean the same thing?

Hint 6:

What if Word and Holy amount to the same?

  • 2
    $\begingroup$ I remark that (aside from modernization of spelling) this is exactly the KJV text of the Johannine Comma, so if there is an actual puzzle here it is there by accident. (Or by divine intervention, or joking around by the KJV translators, but I hope no one will mind my saying that I think those less likely possibilities.) $\endgroup$ – Gareth McCaughan Aug 22 at 21:33
  • $\begingroup$ @GarethMcCaughan In response to the question, Does God have a sense of humor? I was told, “He made you didn’t he?”. Let me assure readers that there are answers and the version and spelling are important. $\endgroup$ – David Aug 22 at 22:23

The two assertions here are (aside from trivial spelling-modernization) exactly the KJV text of the so-called Johannine Comma (a fragment of John's gospel almost universally reckoned to be a later interpolation, which orthodox Christians might regret since it would be the only explicit statement in the Bible of the doctrine of the Trinity). This means that if there is a puzzle here it's the result of coincidence, divine inspiration, or deliberate foolery by the translators of the KJV; I personally think coincidence the most likely explanation. And this in turn means that we shouldn't expect whatever coincidence forms the foundation of the puzzle to be too impressive :-).

So I propose the following: take "there are three ...: X, Y, and Z; and these three are one" to mean that X=Y=Z=1. Likewise, though it's more of a stretch, for "... and these three agree in one". And take X,Y,Z to denote the sums of their letters, the idea being that each letter gets a distinct integer value. (It would be nice for them to be in the range 1..26, but of course requiring the sums to be 1 prevents that.)

There are indeed lots of solutions. If I have translated the question into Mathematica-ese correctly, without the distinct values requirement we would get a simple 10-dimensional space of solutions: let A,B,D,E,F,G,H,I have any integer values at all, and pick two other arbitrary integers that I'll call x and y; then assign L=1+A+2B+C+2H+3x, O=-B-D-H-x, R=1+A+2B-F+2x, S=A+E+F+H-2I-y, T=-A-B-E-H-x, W=F+H, Y=2B+F-G-2H+2I+x+y. Then we just have to pick our values so that these values are distinct. If, e.g., we take A,B,D,E,F,G,H,I,x,y to be 0,1,2,3,4,5,6,7,8,9 then this is the case: we have L=41, O=-17, P=9, R=15, S=-10, T=-18, W=10, Y=14.

If we remove the "...=1" part, which is after all a bit of a stretch for the second verse, then of course there are a lot more solutions; but perhaps now we can get all the values in the range 1..26? (So that we could, if we wanted, fill them out to give an assignment of numbers 1..26 to the whole Latin alphabet.) Yup: take A,B,D,E,F,G,H,I,L,O,P,R,S,T,W,Y to be 23,18,25,17,2,1,3,8,6,10,22,11,16,4,14,7 respectively.

Of course if we really wanted to take this seriously we'd need to do it in Koine Greek, but let's not.

Taking note of the sixth hint,

if in the first sentence we set A=1 ... Z=26 and give up the ghost so that we have just FATHER, WORD and HOLY then these yield totals of 58, 60, 60 respectively. So if instead we take A=2 ... Z=27 then the totals match up (giving 64 in each case, which is appropriate; 64 is both a square and a cube and I'm sure I've seen someone use the way squares form the boundary of a cube as some sort of analogue of the Christian doctrine of the Trinity...). Perhaps this is what @David has in mind?

This doesn't work at all for the second sentence, though:

WATER and BLOOD yield 67 and 48 respectively and are of the same length.

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  • $\begingroup$ I think you are hampered by imagining the solution isn’t that satisfying. And it is not supposed to be so difficult to need Mathematica. But an upvote for the answer assuming it is in some way correct and for the effort put in. $\endgroup$ – David Aug 22 at 22:59
  • $\begingroup$ I am not understanding the answer don’t the L’s cancel? $\endgroup$ – David Aug 22 at 23:23
  • $\begingroup$ I don't think I understand. Cancel where, and why is it a problem? And yes, of course it's possible that there is somehow a more satisfying puzzle here than I was able to find :-). $\endgroup$ – Gareth McCaughan Aug 22 at 23:30
  • $\begingroup$ V jnfa’g guvaxvat nobhg fvkgl sbhe ohg fbzrguvat gbgnyyl qvssrerag. $\endgroup$ – David Aug 24 at 5:33
  • $\begingroup$ V zrna gbgnyyl gur fnzr. $\endgroup$ – David Aug 29 at 17:26

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