This is one possible solving path. The minimum for 1ac is 101 because 2dn can’t start with 0, K = I + 1ac is at least 4 + 101, ie K ≥ 121. From 12ac, O ≥ 121 (because 100 would make 13dn start with 0), and the minimum for s² = u² + O + K is 4² + 121 + 144 = 281, so s > √281 ≈ 16.8. From 21ac, s² + u² ≤ 999, so s < √(999 − 4²) ≈ 31.4. There’s only one square between 16.8 and 31.4, so s = 25 and u, which has to be smaller so that 12ac isn’t negative, is one of {4, 9, 16}.

For 13dn, 25 − B has two digits, so it’s one of {25 − 9 = 16, 25 − 4 = 21}. For 12ac, O = 625 − u² − K, with a maximum of 625 − 4² − 121 = 488, so O is one of {121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484}, but its middle digit has to be 1 or 2 for 13dn, which limits O to {121, 225, 324}. Whichever it is, 13dn = 21 and B = 4, which makes 24ac = 12.

Now 15dn = 29 + y, so y ≤ 99 − 29 = 70 and is one of {9, 16, 36, 49, 64}, so 15dn ends in one of {3, 5, 8}. 21ac = 625 + u² and is either 625 + 9² = 706 or 625 + 16² = 881; but 15dn can’t end in 0, so 21ac = 881 and u = 16, with y in {9, 49} and 15dn in {38, 78}. From 12ac we have O + K = 25² − 16² = 369 and the three possible values for O above make K one of {248, 144, 45}; the only one that’s square is K = 144, making 12ac = O = 225.

From 21dn we have w = 8__ − 625, so w is in the range [800 − 625 = 175, 899 − 625 = 274], ie one of the squares {196, 256} (not 225, which is O), making 21dn one of {821, 881}. But 881 is 21ac, so 21dn = 821 and w = 196. For 22dn we have 1__ = Y − 32, so Y is in the range [100 + 32 = 132, 199 + 32 = 231] and the only unused square in that range is Y = 169, making 22dn = 137. Similarly, for 16ac, 1__ = N − 16, N is in the range [117, 215] and the only available square is N = 121, making 16ac = 105.

From 1dn we have G − n = 25 − 1_, which is in the range [6, 15]. The difference between consecutive squares 64 and 81 is 17, and all differences between higher squares are larger, so G ≤ 64 and is one of {36, 49, 64}, with n < G. The only available squares are {9, 36, 49, 64} and the possible differences ≤ 15 are {64 − 49 = 15, 49 − 36 = 13}. But G − n = 13 would make 1dn = 12, the same as 24ac, so G = 64, n = 49 and 1dn = 10. One of the two values for y above has been taken by n, so y = 9, making 15dn = 38. From 1ac we have I = 144 − 1__, so I < 44 and the only available square is I = 36, making 1ac = 108 and 23dn = 264.

From 2dn, C = 8__ − 242, so C is between 800 − 242 = 558 and 899 − 242 = 657, ie one of {576, 625} and 2dn is one of {818, 867}. From 17dn, P = 5__ − 137 and is between 363 and 462, ie one of {400, 441} with 17dn in {537, 578}. For clue Set 3 the four answer values are 881, one of {818, 867}, one of {537, 578} and 264, and the preamble tells us their sum is a square. The minimum sum possible is 881 + 818 + 537 + 264 = 2500 and the maximum is 881 + 867 + 578 + 264 = 2590; the only square in that range is 2500, so 2dn = 818, C = 576, 17dn = 537 and P = 400.

For Set 1 we now have {108, 225, 23_, 10}, which must sum to a square in the range from 108 + 225 + 230 + 10 = 573 to 108 + 225 + 239 + 10 = 582. There’s only one, so the sum is 576 and 23ac = 233. The clue 23ac gives us L − a = 233 − Y − y = 55. Clue 26dn is L − a − 4, so 26dn = 51. For 28ac we have 4_1 = H − (L − a) − Y, so H = 4_1 + 224 ends in 5 and is in the range 401 + 224 = 625 to 491 _ 224 = 715, and its square root ends in 5 and is in the range √625 = 25 to √715 ≈ 26.7, ie it’s 25, H = 625 and 28ac = 401. Since L and a are squares, L − a can be expressed as f² − g² = 55, which factorises to (f + g)(f − g) = 55, needing either {f + g = 55, f − g = 1} or {f + g = 11, f − g = 5}. The second case gives f = 8 and g = 3 but their squares 64 and 9 are already taken (by G and y); so we want the other case, giving f = 28, g = 27, L = 784 and a = 729.

11dn = t − 928 has two digits, so t is between 10 + 928 = 938 and 99 + 928 = 1027, ie one of {961, 1024} and 11dn is one of {33, 96}. For 27ac we have _1_ = r − t − 85. If t is 1024 then r = _1_ + 1109 and its last two digits are in the range [19, 28]. Its square root is between √(110 + 1109) ≈ 34.9 and √(919 + 1109) ≈ 45.03; of the 11 squares in that range, the ones with 1 or 2 as the penultimate digit are {35² = 1225, 39² = 1521, 45² = 2025}. If t is 1024, for 25ac we have _36 = U − r − 975, so U = _36 + r + 975; then if r is 1225 or 2025, r + 975 ends in 00 and U ends in 36, between 136 + 1225 + 975 = 2336 and 2500 (the maximum letter value allowed), and its square root ends in 4 or 6 and is in the range √2336 ≈ 48.3 to √2500 = 50, but there are no such numbers. If instead r is 1521, we have U = _36 + 2496, which is always > 2500. Therefore t isn’t 1024, t = 961, making 11dn = 33.

Now for 27ac we have r = _1_ + 1046 and its last two digits are in the range [56, 65]. Its square root is from √(110 + 1046) = 34 to √(919 + 1046) ≈ 44.3; of the 11 squares in that range, the ones with matching last digits are {34² = 1156, 42² = 1764}. If r is 1764 then 25ac gives U = _36 + 2676, which can’t be square because it ends in 2. Therefore r = 1156 and 27ac = 110. U = _36 + 2068, ends in 04 and is in the range [2204, 2500]; its square root ends in 2 or 8 and is between √2204 ≈ 46.9 and 50, so it must be 48, making U = 2304 and 25ac = 236.

For 18dn we have c = 1359 − _3. Its square root is between √(1359 − 93) ≈ 35.6 and √(1359 − 13) ≈ 36.7, ie 36, making c = 1296 and 18dn = 63. For 11ac we have l = 2264 − 3_8, so l ends in 6 and its square root ends in 4 or 6 and is between √(2264 − 398) ≈ 43.2 and √(2264 − 308) ≈ 44.2, ie 44, making l = 1936 and 11ac = 328.

For 14ac we have T = __3 + 136, so T is a square ending in 9 in the range [103 + 136 = 239, 993 + 136 = 1129], ie one of {289, 529, 1089} (not 729 = a) and 14ac is one of {153, 393, 953}. For 8dn = 1152 − T, since T ends in 9, 8dn ends in 3, but that’s also the first digit of 14ac, so 14ac = 393, T = 529 and 8dn = 623.

From 20dn we now have E = 3121 − _21, so it ends in 00 and is ≥ 3121 − 921 = 2200, so it can only be E = 2500, making 20dn = 621. Note that we now have all of Set 5 = {623, 63, 621, 137} and their sum is a square, 1444 = 38², as required.

For clue Set 6 we have values {328, 105, 401} and 3dn = __2. We need __2 + 834 to be a square whose root ends in 4 or 6 and is between √(102 + 834) ≈ 30.6 and √(992 + 834) ≈ 42.7, ie one of {34, 36}, so the sum is one of {1156, 1296} and 3dn is one of {322, 462}. That makes 9ac either 12_ or 16_ and from its clue, A = 1008 − (12_ or 16_). For 12_, √A is between √(1008 − 129) ≈ 29.6 and √(1008 − 120) ≈ 29.8, which doesn’t have any integer values. Therefore 3dn = 462 and 9ac needs √A between √(1008 − 169) ≈ 28.97 and √(1008 − 160) ≈ 29.1, ie 29, so A = 841 and 9ac = 167. Now in 3dn we have 462 = 841 + 841 − e − 196, so e = 1024.

For 5dn we have __5 = D + S − 98, so D + S ends in 3, and the squares D and S must end in 4 and 9 (either way round). The smallest unassigned square that matches is 289, so the maximum for D or S is 995 + 98 − 289 = 804, which limits their possible values to {289, 324, 484} and one of them must be 289. Thus 5dn is one of {289 + 324 − 98 = 515, 289 + 484 − 98 = 675}. The second digit of 4ac = W + 176 is 5 or 6, so W’s tens digit is one of {7, 8, 9} and W is one of {81, 676} (not 289, which is D or S; not 576, which is C; not 196, which would make 4ac 372; not 484 or 784, which would make 6dn start with 0). That makes 4ac one of {257, 852}, so 5dn = 515 and D and S are {289, 324} (in either order).

For 10ac we need S − 2o − W to give a two-digit number, so S ≥ 10 + 2×100 + 81 = 291, so it can’t be 289; therefore D = 289 and S = 324. W must be < 324, or 10ac would be negative, so W = 81, making 4ac = 257. Now o = (243 − 10ac)/2 is at most (243 − 10)/2 = 116.5, which only allows o = 100, making 10ac = 43. That completes Set 2 with values {43, 12, 38, 51}, whose sum of 144 is a square, as required.

For Set 4 we have values {515, 33, 821} and 6dn = 74_, so the sum is a square in the range [1369 + 740 = 2109, 1369 + 749 = 2118], ie 2116, with 6dn = 747. The clue for 6dn now gives 747 = R − i + 975, ie i − R = 228. This can be factorised as (f + g)(f − g) = 228, where f = √i, g = √R and both f and g are ≤ 49 (because the maximum square allowed is 2500, which is taken by E). As a product of two integers, 228 is {2×114, 3×76, 4×57, 6×38, 12×19}. The sum of the two factors (f + g) and (f − g) is 2f, even, so they’re either both odd or both even, which leaves only {2×114, 6×38}, but f + g ≤ 97, which rules out 2×114. Therefore f − g = 6 and f + g = 38 and they’re 38/2 ± 3, ie f = 22 and g = 16, making i = 484 and R = 256.

We’ve now assigned all the letter values and solved all the clues, with Set 7 as {257, 167, _3_, 21} and Set 8 as {393, 236, 110, _1_}. The sum for Set 8 is 739 + _1_, which has to be a square with a tens digit of 4 or 5, in the range [739 + 110 = 849, 739 + 919 = 1658]. The only two are {1156, 1444}, but 1444 − 739 = 705 doesn’t fit 19dn, so the sum is 1156 and 19dn = 417. For Set 7, the sum is 445 + _3_, which is a square with a tens digit of 7 or 8, in the range [445 + 130 = 575, 445 + 939 = 1384], ie one of {576, 676, 784, 1089}. 1089 − 445 = 644 doesn’t fit 7dn, but that leaves {131, 231, 339} as possible entries.

To resolve that ambiguity, we need to consider the message from the clue letters, which in ascending numerical order spell “By usInG WoNKY wORDS PiTCH a LAter clUE”. The preamble told us that this “cryptically describes how to obtain an instruction (9,5,6)”, so it’s indicating an anagram (“wonky”) of WORDS PITCH A LATER CLUE, which has 9 + 5 + 6 = 20 letters. Since it’s an instruction, the first word is likely be an imperative verb, like ERADICATE (leaving HLLOPRSTUW), DUPLICATE (ACEHLORRSTW) or CALCULATE (DEHIOPRRSTW). Splitting the remainders into words of 5 and 6 letters, CALCULATE THIRD POWERS looks very convincing. As “the first step in establishing the relationship between the pairs of answers”, let’s try taking the cubes of the numbers in each set, in ascending order:

After some inspection, we can notice that the sum of the smallest and largest cubes in each set is the same as the sum of the middle two, which can’t be a coincidence! For example, in Set 2, 1728 + 132651 = 134379 and 54872 + 79507 = 134379. For Set 7, if the missing cube is 131³ = 2248091 then we have 9261 + 16974593 = 16983854 and 2248091 + 4657463 = 6905554, not equal. With 339³ we have 9261 + 38958219 = 38967480 and 4657463 + 16974593 = 21632056, not a match. But with 231³, 9261 + 38958219 = 16983854 and 4657463 + 16974593 = 16983854 are equal, so 7dn = 231 and the grid is complete.