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 r^{2} − s^{2} must be odd.
The smallest possible value for r^{2} − s^{2} 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 X^{2}, 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 6^{2} + 1^{2} = 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 r^{2} − s^{2} 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, 2^{2} + 7^{2} = 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} + 748^{2}, 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 6^{2} − 1^{2} = 35 (too big),
so it’s 3^{2} − 2^{2} = 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 [47][0257]3_[0257].
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 [678]_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 [47][17]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 = [34]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 = [28]_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[39]_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 = [124578]8[123]6[05];
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 = [56]_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.