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 2^{2} + 2^{2} = 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 = G^{2} 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 = 2G^{2} 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 = 3G^{2} would have three.
For G = 16, 1ac would be 512 and 2dn would be in the range
[16^{2} + 2^{2} = 260, 16^{2} + 8^{2} = 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 = R^{2} + 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 = 2J^{2} 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 = 2N^{2} 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 K^{2} + 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 = C^{2} + P^{2} 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 = P^{2} + 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 = Z^{2} + 4 matches _29,
so Z must end in 5, ie it’s one of {15, 25},
5ac is one of {229, 629}
and 5dn = Z^{2} + 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 = Q^{2} + 40 we have __6 in the grid, so Q ends in 4 or 6.
For 16dn = Q^{2} + 16 we have 5__,
which makes **Q = 24**, **36ac = 616** and **16dn = 592**.

For 13dn = E^{2} + 196 we have 6__, so E is one of {21, 22},
which makes 17dn = E^{2} + 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 = H^{2} + 8 a two-digit entry,
which already contains 57, so **H = 7**.
For 29ac = 2B^{2} 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 = V^{2} + 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, I^{2} + 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 = V^{2} + 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 = O^{2} + 36 we have __7, so O ends in 1 or 9
and the only available value is **O = 9**, making **18ac = 117**.
For 27ac = 2W^{2} 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 = L^{2} + 2 we have 1_1, which fits only **L = 13**,
with **28dn = 171** and **35ac = 178**.
Then for 25dn = Y^{2} + 18 we have _07, so Y ends in 3 or 7
and can only be **Y = 17**, with **25dn = 307**.

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

We have **18dn = 1058** for 2S^{2},
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 − 25^{2} = 576 = 24^{2},
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 − 25^{2} = 495
nor 1120 − 24^{2} = 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 − 25^{2} = 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 − 25^{2} − 21^{2} = 54
isn’t a square,
so there’s no three-square sum starting with 25^{2}.
Similarly, if a three-square sum has a first base of 24
then the second base must be ≥ √((1120 − 24^{2})/2) ≈ 16.5
and the clue-word must batch Q[YFNUEMSQ]_D,
of which only QU_D is a possibility.
That gives 1120 − 24^{2} − 20^{2} = 144 = 12^{2}
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 − 25^{2} = 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 − 24^{2})/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 − 23^{2})/3) ≈ 22.2,
so the largest could only be S,
but there are no four-letter words starting with SS.