## Analogy by Nod

### Puzzle solution process

This is one possible solution path. From Euclid’s formula, the 2rs term must be a multiple of 4 (as either r or s is even), while r2 − s2 must be odd. The smallest possible value for r2 − s2 is 3 (the difference between the first two consecutive squares, 1 and 4), ie with (r, s) = (2, 1), which would produce the triple (3, 4, 5).

In 9dn, the prime P can’t be the multiple of 4, which must be X2, giving X = 2 and P < 4, so P = 3. In 34ac, 3 is the smaller term, so the triple is (3, 4, 5) as above, Q − V = 4 and 9dn = 5n for some value of n.

From 35ac, A > 2V, so V < 101 / 2 = 49.5; thus (V, Q) must be one of (7, 11), (13, 17), (19, 23), (37, 41), (43, 47). In 5dn, R − A must be the (even) 2rs term, so 2V − 3 = (r + s)(r − s) and it evaluates to one of {11, 23, 35, 71, 83}. Four of those are primes, requiring r − s = 1; in the first case, 11 = r + s makes (r, s) = (6, 5) but that gives 2rs = 60, which is > 11; the other primes give even larger values for 2rs, which leaves 2V − 3 = 35, making V = 19, Q = 23 and the triple (3, 4, 5) for 34ac. For (r + s)(r − s) = 35 we must have r + s = 7 and r − s = 5, ie (r, s) = (6, 1), so R − A = 2×6×1 = 12 and 5dn is a multiple of 62 + 12 = 37.

In 35ac we have A − 38 < 57 − R, so A > 38 and A + R < 95, but A + R = 2A + 12, so A < (95 − 12) / 2 = 41.5; the only prime in the range is A = 41, giving R = 53 and making the triple (3, 4, 5), so 35ac = 5n.

For 4dn we have 636 : 41J + 26, where the maximum possible for the r2 − s2 term is 41×101 + 26 = 4167. As 636 = 2rs, rs = 318 = 2×3×53, so the possibilities for (r, s) are (318, 1), (159, 2), (106, 3), (53, 6). All but the last give values > 4167 for the second term, so (r, s) = (53, 6), the triple is (636, 2773, 2845) and J = 67.

For 4ac we have H : 12Y, where 12Y = 2rs and H = (r + s)(r − s). As H is prime, r − s = 1 and r + s = H. As rs = 2×3×Y (with Y being a prime ≥ 5), the possible pairs of factors for (r, s) are (6Y, 1), (3Y, 2), (2Y, 3) and either (Y, 6) or (6, Y), depending on whether Y is greater or less than 6. The first three all give r − s > 1, so Y is either 7 or 5 and H is either 13 or 11, respectively.

For 11ac we have 23Y − D : D − 2, so D > (23Y + 2) / 2. If Y is 7 then D > (23×7 + 2) / 2 = 81.5, and for 161 − D = 2rs to be a multiple of 4 we need D to be 4n + 1 for some n, so D is one of {89, 97, 101}. Alternatively, if Y is 5 then D > (23×5 + 2) / 2 = 58.5, and for 115 − D = 4n we need D = 4n + 3 for some n, so D is one of {59, 71, 79, 83}. We can eliminate (Y, D) = (7, 89) because it gives 72 : 87 for the clue, and the sum of their squares ends in 3 (just using last digits, 22 + 72 = 4 + 9 = 3) and therefore can’t be a square; similarly, (5, 71) gives 44 : 69, with a sum-of-squares ending in 7. From the remaining five options, the only integer triple is (36, 77, 85) with Y = 5 and D = 79. Then H = 11 as above, giving the triples (11, 60, 61) for 4ac and (48, 55, 73) for 31ac.

For 10ac we have 5 : T + 5, so (r + s)(r − s) = 5 and T + 5 = 2rs. The only factors of 5 are 5 and 1, so r + s = 5 and r − s = 1, which means (r, s) = (3, 2), giving T + 5 = 2×3×2 = 12, so T = 7 and the triple is (5, 12, 13). Similarly, for 1ac we have 11 : B − 11 with (r + s)(r − s) = 11 and B − 11 = 2rs, so r + s = 11 and r − s = 1, which means (r, s) = (6, 5), giving B − 11 = 2×6×5 = 60, so B = 71 and the triple is (11, 60, 61).

For 14ac we have 35(LS + N + 19) : 7164L for a 5-digit entry, so L < 99999/7164 ≈ 13.96 and the only available prime is L = 13, which gives 93132 for the second term in 14ac, 12 for the second term in 2dn and 4095 for the second term in 23dn.

For 15dn we have 112 + M : 3239 − 53×O, so M + 53×O < 3127. As 3239 − 53×O must be a multiple of 4, O = 4n + 3 for some n. The minimum value available for M is 17, so 112 + 17 < 3239 − 53×O < 999 (for a 3-digit grid entry), ie O must be between (3239 − 999) / 53 ≈ 42.3 and (3239 − 129) / 53 ≈ 58.7, so O is one of {43, 47}. If O is 43, the second term is 960, with prime factors of {2, 3, 5}, so 112 + M can’t be a multiple of 3 or 5, leaving only {143, 149, 173, 209} as options for the first term; none of these gives an integer for the third term of the triple with 960, so O = 47 instead and the second term is 748.

In 16ac, the first term 287 − M is even, hence it’s a multiple of 4, so M = 4n + 3 for some n. It can’t end with 1 because that would make the sum of squares in 15dn, (112 + M)2 + 7482, end in 3, not a square; that leaves M as one of {43, 59, 83}. The only valid triple for 15dn from those values is (195, 748, 773) with M = 83, and we can enter 15dn = 773.

For 2dn we have 4824 − 79W : 12, so 2rs = 12, rs = 6 and (r, s) is either (6, 1) or (3, 2); the first term can’t be 62 − 12 = 35 (too big), so it’s 32 − 22 = 5, which makes W = 61 and 2dn is a multiple of 13.

For 13ac we have 3G + 7 : 230 + G. As 2rs = 3G + 7 is a multiple of 4, G = 4n + 3 for some n, ie G is one of {31, 43, 59}. As 9dn is a multiple of 5, the grid entry for 13ac must end in 0 or 5 and the third term of the triple ends in 5 (not 0, because it has to be odd). For 31, the terms would be 100 : 261, whose sum of squares ends in 1; for 59, the sum of squares of 184 : 289 ends in 7; so G = 43 and the triple is (136, 273, 305).

In 32ac, the first term 23E + 79 is even, so it’s a multiple of 4, which means E = 4n + 3 for some n; the only such primes left are {31, 59}. If E is 59 then 23dn becomes 828U : 4095 and U < 4095 / 828 ≈ 4.95, but the minimum value available for U is 17. Therefore E = 31, 23dn is 184U : 4095 and U < 4095 / 184 ≈ 22.3, for which the only available value is U = 17. The triple for 23dn is then (3128, 4095, 5153) and 23dn = 5153 can be entered in the grid (because any higher multiples won’t fit).

For 21ac we have 39 : 169 − C. The first term is (r + s)(r − s) and factors as either 39×1 or 13×3. If r + s = 39 and r − s = 1 then (r, s) = (20, 19) and the second term is 2×20×19 = 760, too big; therefore r + s = 13 and r − s = 3, so (r, s) = (8, 5), the second term is 2×8×5 = 80, C = 89 and the triple is (39, 80, 89).

For 27ac we have 15 : 23K − 1245. The first term is (r + s)(r − s) and factors as either 15×1 or 5×3. If r + s = 5 and r − s = 3 then (r, s) = (4, 1) and the second term is 2×4×1 = 8, so K = (8 + 1245) / 23 ≈ 54.5, not an integer. Therefore, r + s = 15, r − s = 1, (r, s) = (8, 7), the second term is 2×8×7 = 112, K = 59 and the triple is (15, 112, 113). This also completes the triples 18ac = (33, 56, 65), 26dn = (1095, 2552, 2777) and 27dn = (48, 55, 73).

For 19dn we have 79S − 2279 : 35, so S < (35 + 2279) / 79 ≈ 29.3. The only value left in range is S = 29 and the triple is (12, 35, 37). The triple for 33ac is now complete as (385, 552, 673) and we can write in 33ac = 673 (because higher multiples won’t fit).

For 29ac we now have 61N − 2209 : 55, where 55 = (r + s)(r − s) factors as 55×1 or 11×5. If r + s = 55 and r − s = 1 then (r, s) = (28, 27) and the first term is 2×28×27 = 1512, too big. Therefore r + s = 11, r − s = 5, (r, s) = (8, 3), the first term is 2×8×3 = 48, N = 37 and the triple is (48, 55, 73). This also completes the triples 12ac = (39, 80, 89), 14ac = (15155, 93132, 94357), 1dn = (420, 851, 949), 3dn = (57, 176, 185), 6dn = (4325, 14652, 15277), 8dn = (11448, 44215, 45673) and 28dn = (252, 275, 373), and 14ac = 94357 and 1dn = 949 can be entered in the grid.

In 20ac, the first term is 19Z − 1891, so Z > 1891 / 19 ≈ 99.5, which means Z = 101, giving the triple (28, 45, 53). This also completes 32ac = (792, 1855, 2017).

For 24ac we have 1266 − 13F : 12, so F > (1266 − 12) / 13 ≈ 96.5, which means F = 97, giving the triple (5, 12, 13) and also completing 7ac = (52, 165, 173) and 25dn = (364, 627, 725). The only prime left is I = 73, which completes the remaining triples, 16ac = (204, 253, 325), 22ac = (215, 912, 937), 7dn = (20, 21, 29), 14dn = (20, 21, 29), 17dn = (341, 420, 541), 20dn = (165, 532, 557), 21dn = (2848, 7665, 8177) and 30dn = (7, 24, 25).

Now we can complete the grid. In addition to the numbers already entered, we can enter 17dn = 541 immediately, as higher multiples have more than 3 digits.

8dn is a multiple of 45673, ie one of {45673, 91346}. From the grid, 16ac ends in 5, so as a multiple of 325 its last two digits must be either 25 or 75, so 8dn = 45673. Then 21ac = 89n = _34, with the last digit of n being 6; 89×16 = 1424 is too big, so n = 6 and 21ac = 534. Then 21dn = 8177n = 5____, where the only multiple that fits is 21dn = 57239 (with n = 7).

28dn = 373n is one of {373, 746}, 25dn = 725n ends in one of {00, 25, 50, 75}, and 30dn = 25n ends in one of {00, 25, 50, 75}, so 32ac = 2017n matches some combination of 3_. If it starts with 7 then n is between 70300 / 2017 ≈ 34.9 and 77397 / 2017 ≈ 38.4, but none of the resultant multiples {70595, 72612, 74629, 76646} fit; therefore 28dn = 746, and for 32ac n is between 40300 / 2017 ≈ 19.98 and 47397 / 2017 ≈ 23.5, for which only {40340, 42357} fit.

26dn = 2777n is one of {2777, 5554, 8331} but from 32ac the third digit is 4 or 5, so 26dn = 5554 and 32ac = 42357. Then 30dn ends in 75, making 35ac = 945, and 25dn ends in 25 and is one of {3625, 6525, 9425}. So 24ac = 13n is one of {33751, 36751, 39751}, of which only 24ac = 36751 is a multiple of 13, forcing 25dn = 6525. Now 29ac = 73n = 525_, which can only be 29ac = 5256, completing 30dn = 675.

14dn = 29n = 9__ is one of {928, 957, 986}, so 22ac = 937n matches _5_; the only multiples between 6000 and 8999 are {6559, 7496, 8433}, so 22ac = 6559 and 14dn = 986. Then 18ac = 65n starts with 8 and the only multiple in range is 18ac = 845.

27dn = 73n ends in 6, so n ends in 2 and is between 116 / 73 ≈ 1.6 and 996 / 73 ≈ 13.6, ie n is 2 or 12 and 27dn is one of {146, 876}.

19dn = 37n = 45__7, so n ends in 1 and is between 45007 / 37 ≈ 1216.4 and 45997 / 37 = 1243.2, ie n is one of {1221, 1231, 1241} and 19dn is one of {45177, 45547, 45917}. 27ac = 113n ends in 7, so n ends in 9; the entry matches one of {111_7, 151_7, 191_7, 811_7, 851_7, 891_7}, for which the corresponding ranges for n are (98.3, 99.1), (133.7, 134.5), (169.1, 169.9), (717.8, 718.6), (753.2, 753.96), (788.6, 789.4), ie n is one of {99, 789} and 27ac is one of {11187, 89157}, which limits 19dn to {45177, 45917}. Then 31ac = 73n matches 5_ one of {415_, 475_, 715_, 775_}, for which the corresponding ranges for n are (56.8, 56.97), (65.1, 65.2), (97.95, 98.1), (106.2, 106.3), ie n = 98 and 31ac = 7154, which makes 27dn = 876, 19dn = 45917 and 27ac = 89157.

20ac = 53n = _7_; of the 3-digit multiples of 53, the only ones with 7 in the middle are {371, 477}. Then 20dn = 557n = 954_, so n is either in the range (70.99, 71.004) or in the range (88.94, 88.96), ie n = 71, 20dn = 39547 (completing 34ac = 765) and 20ac = 371.

Now 6dn = 15277n ends in 1, so n = 3 and 6dn = 45831. 16ac = 325n = 3_75, so n = 11 and 16ac = 3575. 7ac = 173n = _4_; the only multiple that fits is 7ac = 346 with n = 2. 11ac = 85n = 5_5_; for multiples between 5050 and 5959, n is in the range (59.4, 70.1) and the only multiples that fit are {5355, 5950}.

4dn = 2845n ends in 5, so n is one of {1, 3} and 4dn is one of {2845, 8535}. 4ac = 61n = _4, so n ends in 4 and is in the range (3.5, 4.8) or (13.3, 14.7), ie n is one of {4, 14} and 4ac is one of {244, 854}. Then 5dn = 37n starts with 4 or 5, so n is in the range (10.8, 16.2), ie one of {11, 12, 13, 14, 15, 16} and 5dn is one of {407, 444, 481, 518, 555, 592}.

7dn = 29n = 3_5, so n ends in 5 and is in the range (113.97, 137.8), ie n is one of {115, 125, 135} and 7dn is one of {3335, 3625, 3915}. 13ac = 305n = 86; the only value that is a multiple of 305 is 13ac = 28365 with n = 93, which makes 7dn = 3335 and 11ac = 5355, completing 9dn = 655. Now 5dn ends in 2, so 5dn = 592, 4ac = 854, 4dn = 8535.

1ac = 61n = 9__, so n is between 911 / 61 ≈ 14.9 and 999 / 61 ≈ 16.4, ie n is one of {15, 16} and the entry is one of {915, 976}. 12ac = 89n = 9__3, so n ends in 7 and is in the range (101.2, 112.3), ie n = 107 and 12ac = 9523.

3dn = 185n = _235; the only multiple that fits is 3dn = 61235 with n = 331, which makes 1ac = 976. Then 2dn = 13n = 7_54; the only multiple that fits is 2dn = 7254 with n = 558, which completes 10ac = 42159 and the grid.