This is one possible solving path. Since the clues are of the form (2p + 1)² + (2q + 1)² + (2r + 1)² = 16s + 3, all four terms are odd, which allows many grid cells to be identified as odd digits (shaded orange here). Also, for A − B (clue I) and F − e (clue IV) to be odd, A and F must be even (their last digits shaded blue). As s in the equations is a triangular number, for every clue the (sum − 3)/16 must be triangular (and an integer).

The squares of odd numbers ending in {1, 3, 5, 7, 9} end in {1, 9, 5, 9, 1} respectively. In clue VII the sum 2[k]63 ends in 3, so the three square terms must end in either (1, 1, 1) or (5, 9, 9) in some order. For (1, 1, 1), all of {n, m, K} must end in 1 or 9; m and K share the same first digit and the clue tells us m < K, so they would be _1 and _9 respectively, making n either 91 or 99, but then n < m is impossible. So the squares end in (5, 9, 9), one of {n, m, K} ends in 5 and each of the other two ends in 3 or 7. If n is _5 then m is _3 and K is _7, making n = 75, which in turn pushes m and K up to {83, 87} or {93, 97}; but 75² + 83² + 87² = 20083 and 75² + 93² + 97² = 23683, neither of which matches the clue’s sum of 2__63, so n doesn’t end in 5. If K ends in 5 then n is one of {53, 57} and for n < m < K we need m = K − 2 to be one of {63, 73, 83, 93}; choosing maximum values, the equation is 57² + 93² + 95² = 20923, but to contain k the sum must be ≥ 21163. Therefore m ends in 5, K = m + 2 (to end in 7) and n is one of {73, 77}. Of the combinations {73² + 75² + 77² = 16883, 73² + 85² + 87² = 20123, 73² + 95² + 97² = 23763, 77² + 85² + 87² = 20723, 77² + 95² + 97² = 24363}, the only sums ≥ 21163 are with m = 95 and K = 97, namely {23763, 24363}. We need the sum to match 16s + 3, ie having a remainder of 3 when divided by 16; 24363 = 16×1522 + 11 doesn’t work, so we’re left with 23763 = 16×1485 + 3, making n = 73 and k = 37.

In clue VIII we now have N = _5, starting with an odd digit, and N < 37, so it’s one of {15, 35}. H starts with 3, so for clue II the sum is at least 3003, but if N is 15 then the maximum sum possible is 11² + 13² + 15² = 515, too small; therefore N = 35 and for clue VIII we have 2594 + h² = 4[G]. L is _7, so it’s one of {17, 57, 77}, as 37 and 97 are already used, which makes h > 37 end in one of {1, 5, 7}. The maximum for h is √(4999 − 2594) ≈ 49.04, so h is one of {41, 45, 47}. The corresponding sums for clue VIII are {4275, 4619, 4803} and the values for (sum − 3)/16 are {267, 288.5, 300}. Testing for triangularity, √(2×267) ≈ 23.1, but the 23rd triangular number is 23×24/2 = 276; the only triangular option is 300 = 24×25/2, so the sum is 4803, G = 803 and h = 47, making L = 77.

In clue VI, the middle term needs F − 73 ≤ 75, so F ≤ 148 and the sum is in the range [10051, 14851]. For clue V we need b < D < a; b and D share the same last digit, so b ≤ D − 10; D’s (odd) first digit is a’s second, so the digits of a can’t be in ascending order. The sum d is ≥ 11001 (from the grid), but if a starts with 6 or lower, the maximum is 49² + 59² + 65² = 10107, too small. Omitting values already used, a is one of {75, 83, 85, 87, 93, 99} and the maximum for the sum d is 83² + 93² + 99² = 25339 (D can’t be 95 or 97, which are already used). That means B starts with 1 or 2. If it’s 19 then the maximum for clue II is 19² + 33² + 35² = 2675, but we need it to be ≥ 3003; therefore B and d start with 2, and M is in the range [23, 33]. If a < 90 then the maximum sum for clue V is 69² + 79² + 87² = 18571, too small for d, so a starts with 9; it can’t be 91 because D’s first digit is at least 3 (so that b can be smaller). If a is 93 then the maximum sum is 29² + 39² + 93² = 11011, too small; that only leaves a = 99, making D one of {91, 93}.

If D and b end with 1 then the sum d ends with 3; if they end in 3 then d ends in 9; that limits M to one of {23, 29, 33}. If M is 29 then the maximum for clue II is 27² + 29² + 35² = 2795, too small. Therefore M = 33 and d ends in 3, so D = 91. The minimum for B is now √(3003 − 33² − 35²) ≈ 26.2, so B is one of {27, 29}, giving corresponding sums of {3043, 3155} and (sum − 3)/16 values in {190, 197}. Testing for triangularity, √(2×197) ≈ 19.8, so we want the 19th triangular number, 19×20/2 = 190, the other one, so [H]3 = 3043, H = 304 and B = 27.

In the grid, d = 2__43 and is in the range [21043, 29943], with its associated triangular number t = (d − 3)/16 between 1315 and 1871.25, with a triangular root in the range [51, 61]. But t ends in 0 or 5 and 2t = n(n + 1) ends in 0, so the root n must end in one of {4, 5, 9, 0}, restricting the roots to {54, 55, 59, 60}. The corresponding values for t are {1485, 1540, 1770, 1830}, making d one of {23763, 24643, 28323, 29283} and the only one that fits the grid is d = 24643, with b = 81.

For clue III, e² + C² + f² = 14[E], E is now in the range [411, 499], giving limits for the sum of 14411 and 14499. The associated triangular number is between 900.5 and 906; √(2×901) ≈ 42.4 and √(2×906) ≈ 42.6, so the triangular has to be 42, giving the triangular number 903 and the sum of 14451, which makes E = 451 and e ends in 1. To give a sum ending in 1, the three squares must end in either (1, 1, 9) or (1, 5, 5), and we already know e² ends in 1. If the other two squares both end in 5 then C and f also end in 5, ie f is 55; but then the maximum sum would be 41² + 45² + 55² = 6731, too small. So either C² or f² ends in 1 and the other ends in 9; that is, either C or f ends in 1 or 9 and the other ends in 3 or 7. C can’t end in 1 (because C > e), so f is one of {31, 39, 71, 79, 93} (not 97, already used). If f is 71, the maximum sum is 61² + 67² + 71² = 13251, too small. If f is 93 then C ending in 9 is e + 8, giving e² + (e + 8)² + 93² = 14451, ie e² + 8e − 2869 = 0, for which the quadratic formula gives the roots {49.7, -57.7} (approximately), not integers. What’s left is f = 79, and the quadratic equation e² + (e + 6)² + 79² = 14451 has roots {61, -67}, so e = 61 and C = 67.

For clue I we now have (A − 27)² + 77² + 83² = 158_3, so (A − 27) is between √(15803 − 12818) ≈ 54.6 and √(15893 − 12818) ≈ 55.5, so A − 27 = 55, ie A = 82, the sum is 15843 and g = 5843. For clue VI we have 59² + (F − 73)² + 77² = [F]51. As the sum ends in 1, the three squares must end in (1, 1, 9) or (1, 5, 5), but we know the first and third ones end in 1 and 9 respectively, so (F − 73)² must end in 1, with (F − 73) ending in 1 or 9; the possibilities between 59 and 77 are {61, 69, 71}, giving F as one of {134, 142, 144} and the corresponding sum in {13131, 14171, 14451}. The only one that matches the grid entry is 14451 with F = 144. Clue IV now becomes 67² + 83² + 95² = 20403, so c = 2040 and the grid is full. The eight equations we have are:

55² + 77² + 83² = 15843 27² + 33² + 35² = 3043 61² + 67² + 79² = 14451 67² + 83² + 95² = 20403 81² + 91² + 99² = 24643 59² + 71² + 77² = 14451 73² + 95² + 97² = 23763 35² + 37² + 47² = 4803

For each equation’s square roots (2p + 1), (2q + 1) and (2r + 1), each of the triangular numbers t = p(p + 1)/2, u = q(q + 1)/2, v = r(r + 1)/2 is the sum of a pair of numbers in a thematic triple (a, b, c), let’s say t = a + b, u = a + c, v = b + c. Substituting c = v − b into the second equation gives u = a + v − b, then substituting b = a + v − u into the first one gives t = 2a + v − u, ie a = (t + u − v)/2. Then we can get b = t − a and c = v − b.

For example, in clue I we have p = (55 − 1)/2 = 27, q = (77 − 1)/2 = 38 and r = (83 − 1)/2 = 41 for the roots of the triangular numbers {27×28/2 = 378, 38×39/2 = 741, 41×42/2 = 861}, from which we can derive the thematic triple {a = (378 + 741 − 861)/2 = 129, b = 378 − 129 = 249, c = 861 − 249 = 612}. As a check, the sum of all three is 129 + 249 + 612 = 990 = 44×45/2, a triangular number. The full set of triangular numbers derived from the clues and their associated thematic triples is:

{378, 741, 861} → {129, 249, 612} {91, 136, 153} → {37, 54, 99} {465, 561, 780} → {123, 342, 438} {561, 861, 1128} → {147, 414, 714} {820, 1035, 1225} → {315, 505, 720} {435, 630, 741} → {162, 273, 468} {666, 1128, 1176} → {309, 357, 819} {153, 171, 276} → {24, 129, 147}

The numbers that appear more than once are 129 and 147; their sum is 276 = 23×24/2, the 23rd triangular number. (Adding both occurrences of each of them would give 552, but that doesn’t have the thematic property of being triangular.)