Clue-by-four by gwizardry

Puzzle solution process

This is one possible solution path. We can see that all 26 letters are used in the clues, and one of them must be 0; the only letter that never appears to the left of a different letter is D, so D = 0.

The letters that appear in clues of the form x x D D are {B, F, G, J, N, S, W}, which must all be > 2, because 22 + 22 = 8 is too short for a grid entry. All the other letters appear to the left of a letter other than D and therefore must be > 1, apart from T, so T = 1.

The letters that appear to the left of a G are {A, C, E, I, K, Q, R, V}, and 17ac has Y > A, so G is less than all nine of those letters, ie G ≤ 16. As 33dn = G2 has at least two digits, G ≥ 4. The letters appearing to the left of X are {C, G, H, I, J, K, N, P, Q, R, Z}, and we have M > P (14ac), O > J (18ac), A > G (24ac), Y > A, U > J (10dn), E > C (13dn) and V > I (19dn), so X is less than all 17 of those, ie X ≤ 8.

There’s no 1dn, so 1ac must start in an unchecked cell in the top-left corner. Because there are no consecutive unchecked cells, the second cell of 1ac must be the first cell of 2dn. As G is in the range [4, 16], 1ac = 2G2 is in the range [32, 512]. We can eliminate G = 5 or 10 because they would give 50 or 200 for 1ac, making 2dn start with 0; for 6 and 7, 1ac would have two digits but the symmetrically opposite 37ac = 3G2 would have three. For G = 16, 1ac would be 512 and 2dn would be in the range [162 + 22 = 260, 162 + 82 = 320], but no numbers in that range start with 1. Similarly, for G = 15, 1ac = 450 but 2dn in the range [229, 289] clashes; for G = 14, 1ac = 392 clashes with 2dn in [200, 260]; for G = 13, 1ac = 338 clashes with 2dn in [173, 233]; and so on, with only G = 4, 1ac = 32 allowing a match. If X is then 3, 2dn would be 25, but as that’s a square its clue would have to be 5 0 0 0, not 4 3 0 0; so X = 2 and 2dn = 20.

We can add bars at the end of 1ac and 2dn, and symmetrically opposite we have 34dn = 24 at the bottom of column 7 and 37ac = 48 at the end of the bottom row. The second cell in row 2 contains 0, so it can’t be the start of an across entry (and it must be checked because it’s not on the perimeter), so there’s an across entry at the start of row 2, which means that row 7 ends with an entry including the 2 at the start of 34dn. As there’s no 34ac, that entry must be 33ac. Since 33dn = 16, 33ac = R2 + 20 must be 12_, for which only 33ac = 120 with R = 10 fits. As 33ac is the fourth-to-last across entry, its opposite number must be 8ac at the start of row 2, so we can enter 8ac = 102 and 8dn = 17.

We have both 3ac and 3dn, so they must start in the cell immediately after 1ac, 3dn = 2J2 being opposite 33dn and matching _2. The only value that fits is 3dn = 72 with J = 6 (32 being already used at 1ac). Now 3ac = 2N2 starts with 7; the only two-digit value (72) is already taken, so 3ac = 722 with N = 19, which places 36ac as the three-digit entry opposite, ending in 6. The top row now has only three free cells for 5ac and the starts of 5dn, 6dn and 7dn, so the latter three start in the last three columns and 5ac is a fully-checked three-digit entry, as is 35ac opposite; 4dn must start with one of the 2s in 3ac, the other one being unchecked.

Row 7 must have at least one across entry in the five cells before 33ac (or there would be internal unchecked cells) and no more than two. If there’s only one, namely 32ac, then 31ac would have to be in row 6, but there’s a down entry in the last column (opposite 8dn), which would have to be numbered between 31 and 32; so 31ac and 32ac are both in row 7 and the entry in the last column has to be 30dn (as there’s no 29dn). It is K2 + 4 and matches _0, for which only 30dn = 80 fits, with K = 8.

The clue for 7dn tells us K > P, and the only available values for P are {3, 5, 7}. It can’t be 3 because that would give 17 for 11ac and 17 is already used at 8dn.

9ac must start in the cell immediately after 8ac (or it would be unchecked). As there’s no 9dn, the first digit of 9ac must be the second digit of 4dn (in column 4) and the last digit of 3ac is unchecked. As the second cell of 9ac can’t be unchecked, it must be the start of 10dn. 4dn = C2 + P2 which starts with 2. If P is 5, C has to be one of {14, 15, 16} to make 4dn one of {221, 250, 281} respectively; if P is 7, C is one of {13, 14, 15} and 4dn is {218, 245, 274} respectively. Either way, 4dn has three digits and the opposite entry must be a three-digit 28dn. There are two combinations for (P, C, 4dn, 9ac) that fit together, (5, 14, 221, 216) and (7, 13, 218, 189). Either way, 9ac has three digits, so 11ac can only have two, and opposite we have 31ac/31dn starting in the first column and 32ac in the third.

Now 7dn = P2 + 68 has two digits; if P is 7 then 7dn would be 117, so P = 5, 7dn = 93, C = 14, 4dn = 221, 9ac = 216, 11ac = 33, 6dn = 236, 31ac = 30 and 31dn = 31.

Now 5ac = Z2 + 4 matches _29, so Z must end in 5, ie it’s one of {15, 25}, 5ac is one of {229, 629} and 5dn = Z2 + 38 is one of {263, 663}.

In row 3, 12ac = 76 can’t start in the second cell, because there’s no 12dn and the cell would be unchecked, so it starts in the first cell with the 7 from 8dn. Then 13dn starts in the second cell, and 14ac/14dn in the next.

In row 6, the two three-digit down entries finishing in 35ac must be 25dn and 26dn, so 27ac can only start in the cell between 26 and 28. As there’s no 25ac, the row must start with a three-digit 24ac. Symmetry puts 15ac = 365 opposite at the end of row 3, with 16dn in the last column. As we have both 17ac and 17dn, they must start in the first cell of row 4, and 16dn and 17dn are three-digit opposite entries. For 36ac = Q2 + 40 we have __6 in the grid, so Q ends in 4 or 6. For 16dn = Q2 + 16 we have 5__, which makes Q = 24, 36ac = 616 and 16dn = 592.

For 13dn = E2 + 196 we have 6__, so E is one of {21, 22}, which makes 17dn = E2 + 16 {457, 500} respectively. But 500 would make 24ac start with 0, so E = 21, 17dn = 457 and 13dn = 637.

As 20ac and 21ac are the middle two across entries, they must be symmetrical opposites, with 20ac in row 4 and 21ac in row 5. The penultimate cell in row 4 must be the start of a down entry (or it would be unchecked), so that’s where 20ac and 20dn start. That makes 21ac = H2 + 8 a two-digit entry, which already contains 57, so H = 7. For 29ac = 2B2 we have _8 in the grid, which can now be satisfied only by B = 3, giving 29ac = 18 and 20ac = 19. Opposite 17ac we have 23ac = V2 + 17 ending in the last cell of row 5, which contains a 2, so V ends in 5 and is one of {15, 25} and 23ac is one of {242, 642}. Either way, I2 + 20 = 20dn = 141, so I = 11.

We now have all the grid bars in place and can add the remaining entry numbers for 18ac/18dn, 19dn and 22ac = 802.

For 19dn = V2 + 137 is one of {362, 762}. Either way, 23ac = 642, which makes V = 25 and 19dn = 762, leaving Z = 15, 5ac = 229 and 5dn = 263.

For 18ac = O2 + 36 we have __7, so O ends in 1 or 9 and the only available value is O = 9, making 18ac = 117. For 27ac = 2W2 we have __2, so W ends in one of {1, 4, 6, 9} and the only available value is W = 16, making 27ac = 512. For 28dn = L2 + 2 we have 1_1, which fits only L = 13, with 28dn = 171 and 35ac = 178. Then for 25dn = Y2 + 18 we have _07, so Y ends in 3 or 7 and can only be Y = 17, with 25dn = 307.

Now for 24ac = A2 + 592 we have 73_, so A = 12, 24ac = 736 and 17ac = 433. Then 26dn = 2F2 is 6_8, so F = 18 and 26dn = 648. And then 32ac = M2 + 3 is 4_7, so M = 22, 32ac = 487 and 14ac = 510.

We have 18dn = 1058 for 2S2, so S = 23, confirming 14dn = 538. That only leaves U = 20, confirming 10dn = 1012. The grid and all letter values are now all established.

The central four cells contain 1, 1, 0, 2, from which the possible four-digit numbers are {1120, 1201, 2011}. None of them is itself a square, so next we check whether they have two-square sums.

For any such n, the two squares must be on opposites sides of (and equidistant from) n/2, so one of the bases is above √(n/2) and the other is below (or they’re both equal to it). For 2011, √(2011/2) ≈ 31.7, but our bases must be ≤ 25, so 2011 has no two-square sum that we can use. For 1201, √(1201/2) ≈ 24.5, so the larger base can only be 25; in fact, 1201 − 252 = 576 = 242, so the clue for 1201 would be 25 24 0 0 = V Q D D, not a word. For 1120, √(1120/2) ≈ 23.7, but neither 1120 − 252 = 495 nor 1120 − 242 = 544 is a square, so 1120 doesn’t have a usable two-square sum.

For any n that is the sum of three squares, the squares must be spread around n/3: either all three equal to n/3, or one above, one below and the other between them. For 2011, √(2011/3) ≈ 25.9, which doesn’t allow one of our bases to be greater (or equal), so 2011 doesn’t have a (usable) three-square sum.

For 1120, √(1120/3) ≈ 19.3, so the first base of a three-square sum would have to be ≥ 20, ie one of {U, E, M, S, Q, V} (in ascending order). If it’s 25 then the remainder 1120 − 252 = 495 would have to be a two-square sum, with bases either side of √(495/2) ≈ 15.7, the larger one being one of {W, Y, F, N, U, E, M, S, Q, V}, ie the clue-word would have to match V[WYFNUEMSQV]_D, of which only VE_D could be a common word (VEND or maybe VELD). But 1120 − 252 − 212 = 54 isn’t a square, so there’s no three-square sum starting with 252. Similarly, if a three-square sum has a first base of 24 then the second base must be ≥ √((1120 − 242)/2) ≈ 16.5 and the clue-word must batch Q[YFNUEMSQ]_D, of which only QU_D is a possibility. That gives 1120 − 242 − 202 = 144 = 122 and 12 is A, so the clue-word for 1120 could be QUAD. In fact, it has to be, because any further calculations would only find three-sums with a smaller first base, or four-sums.

For completeness, we should see whether 2011 has a four-square sum that produces a more thematically appropriate word than QUAD. The bases would have to be spread around √(2011/4) ≈ 22.4, so the first would be one of {S, Q, V}. For V, the next base would be the first for a three-square sum to 2011 − 252 = 1386, whose bases are spread around √(1386/3) ≈ 21.5, ie one of {M, S, Q, V}, but there are no four-letter words starting with VM, VS, VQ or VV. For Q, the bases of the remainder would be spread around √((2011 − 242)/3) ≈ 21.9, so the largest would be one of {M, S, Q}, but there are no four-letter words starting with QM, QS or QQ. Finally, for S, the bases of the remainder would be spread around √((2011 − 232)/3) ≈ 22.2, so the largest could only be S, but there are no four-letter words starting with SS.

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