## Pairs by Elap

### Puzzle explanation

The conversion process consists of squaring a number, inserting a gap to split the result into two squares, taking the square roots of those numbers and then removing the gap to join the roots together. All grid entries are matched up in this way except for 1771 at 17dn, which leads to 5621.

### Puzzle solution process

This is one possible solution path. It will be helpful to list the possible values consisting of one or two squares, up to two digits, as follows: 1, 4, 9, 11, 14, 16, 19, 25, 36, 41, 44, 49, 64, 81, 91, 94, 99. There are 55 three-digit values, from 100 to 981, but we’ll avoid the drudgery of listing them all.

From 17dn, G(G + 1) + P has 4 digits, so the minimum value for G is close to √(1000 − 99) = √901. As 252 = 625, G > 25 and the lowest permissible value from the list above is 36. As 19ac = Y2 + G has 2 digits, the maximum for Y2 is 99 − 36 = 63, so Y ≤ 7 and the only permissible values are 1 and 4. Y can’t be 1 because that would make the entries 12ac = IY + I + Y and 7dn = IY + Y2 + I the same, so Y = 4.

As 13dn = I3 + ET has 3 digits, I < 10. I can’t be 1 because that would make 23dn = I + Y = 5, too short, so I = 9 and P = 49 (from 12ac). This also gives 7dn = 61 and 23dn = 13 in the grid.

10ac = 4E2 + X has 4 digits, so E < √(10000/4) = 50, which means 17ac < 50 and 17dn < 5000. The maximum for 17dn = G(G + 1) + 49 is now 4999, so the maximum for G is close to √(4999 − 49) ≈ 70.4, which restricts G to one of {36, 41, 44, 49, 64}. Trying these values in 17dn and 19ac = 16 + G, they all give a clash in the second cell except for {41, 44}.

14dn = S2 + X has 3 digits, so S < √1000 ≈ 31.6 and the available values are {11, 14, 16, 19, 25} (not 1, because that would make 25ac = S + X and 14dn = SS + X the same).

From 22dn, G = L + S, for which the only possibilities are G = 41 and {L, S} = {16, 25} in either order, or G = 44 and {L, S} = {19, 25} in either order. If G is 44 then 17dn is 2029, forcing 17ac = E to be 25, but 25 is already taken by either L or S. Therefore 22dn = G = 41, 17dn = 1771, 19ac = 57 and {L, S} = {16, 25}.

17ac = E is now one of {11, 14, 19} (not 16, which is already taken by L or S), so 24dn = 16 + E is one of {27, 30, 35} which means 23ac = 169 + T ends in 2 or 3, so T ends in 3 or 4. But there are no permissible values ending in 3, so T ends in 4 and must be at most 193 − 169 = 24; the only available possibility is T = 14, giving 23ac = 183. Then 24dn = 35 and E = 19, giving 17ac = 19, 13dn = 995 and 21dn = 285.

15ac = M + 2X has 3 digits, so X < 500, which means 10ac = 1444 + X < 1944 and we can enter the 1. Then 10dn = 2L + N (3 digits) starts with 1, so 20dn = N (3 digits) must also start with 1. Then we have 20ac = L2 + Z = 152_ and the maximum permissible for Z is 981, so the minimum for L2 is 1520 − 981 = 539; 16 is too small for that, so L = 25 and S = 16 and we can enter 1dn = 43. Z is now in the range 1520 − 625 = 895 to 1529 − 625 = 904, so the only permissible value is Z = 900 and 20ac = 1525.

Now 14dn = 256 + X ends in 5, so X ends in 9, which means 25ac = 16 + X is _15 (the 1 was already in the grid). Therefore X ends in 99, and its only permissible values are {99, 499}.

From the grid, 20dn = N is 1_5. There are no squares that fit (as 52 = 25 and 152 = 225), so the only permissible value is N = 125, which makes 10dn = 175. Then 15ac = M + 2X is 7__, so X can’t be 499 and we have X = 99, 10ac = 1543, 25ac = 115 and 14dn = 355.

For 6dn we have 10J + 41, so it ends in 1; that completes 915 as the entry for 18ac = B + 914, so B = 1 and we can enter 8ac = 305. Then 3dn = F + 29 is _5, so F is _6. As 16 is already assigned to S, the only permissible value is F = 36 and we can enter 9ac = 721 and 3dn = 65.

For 11dn = 4D + 99 the grid gives 4_5, so D must be one of {79, 84, 89, 94, 99}. The first 3 values aren’t valid and 99 is already assigned to X, so D = 94 and 11dn = 475.

16dn = 49A + 125 has 4 digits, so the minimum for A is (1000 − 125)/49 ≈ 17.9. 22ac = 9A + R is 4__, so the maximum value for A is (499 − 11)/9 ≈ 54.2. The only unassigned permissible value in that range is A = 44, so we can enter 16dn = 2281. That completes 15ac = M + 198 as 727, so M = 529 and we can enter 5dn = 1249. For 22ac = 396 + R we have 48_, so R is in the range [84, 93]; the only permissible value in that range is R = 91, and we can enter 22ac = 487 and 4dn = 223.

We now have 4ac = J + 99 as 21_, so J is in the range [111, 120]. There are no squares in that range (as 102 = 100 and 112 = 121), so the only permissible value is J = 116 and we can enter 4ac = 215 and 6dn = 1201. That leaves 2dn = 616 + U as _05, so U is one of {89, 189, 289}, of which the only permissible value is U = 289 and 2dn = 905. The grid is now complete.

Arranging the letters in numeric order, we have BY ITSELF GAP RDXJNUMZ, suggesting that the first step in the conversion process is multiplying an entry by itself, the second step is inserting a gap and the fourth step is removing a gap.

Squaring the grid entries, we can see that several of them can be split (by inserting a gap) into two smaller squares, like the letter values. For example, (23dn) 132 = 16 9, (17ac) 192 = 36 1, (24dn) 352 = 1 225, (22dn) 412 = 16 81 and (6dn) 12012 = 144 2401. As the fourth step involves removing a gap, the third step presumably leaves us with two numbers. A natural thing to do with the squares is to take their roots, and it turns out that joining the roots together gives another grid entry in most cases, as below. (The rest of the clue letters, RDX JNUMZ, might be taken as hints for “radix” and “join numbers”, but probably only in a subliminal way.)

Entry Value Squared, split Roots joined Entry
23dn 13 16 9 43 1dn
17ac 19 36 1 61 7dn
24dn 35 1 225 115 25ac
22dn 41 16 81 49 12ac
19ac 57 324 9 183 23ac
3dn 65 4 225 215 4ac
20dn 125 1 5625 175 10dn
4dn 223 49 729 727 15ac
21dn 285 81 225 915 18ac
8ac 305 9 3025 355 14dn
11dn 475 225 625 1525 20ac
22ac 487 23716 9 1543 10ac
9ac 721 51984 1 2281 16dn
2dn 905 81 9025 995 13dn
6dn 1201 144 2401 1249 5dn
17dn 1771 3136 441 5621

The unmatched entry is 17dn = 1771, which the process converts to 5621, to be written below the grid.