Overview:
- fill most cages
- refine

Fill most cages
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When the sum's 1 more thena the product P, you've got a [1P] pair.
They all line up, so it's 1-2 above 3/1 next to 4/1.
Bottom right is a [23], as yet unresolved, leaving a [56] pair.

Refine
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Sudoku slog, alas.

Column 5's white kropke must have the 5 on it
row 4's [456] triple forces 2 into r4c4 and 1 into r4c5
this restricts row 2's kropke to [23]o[14], and leaves an X-wing of 1s in r[12]c[16]
row 6's kropke can't have a 6 on it, as that woudl breal r6c5, and now must have teh box's 4
=> box 1's 6 is in r[12]c[3]

r5c2 can't be 5 as that places 2 in r6c1 and 5 in r6c5, and there's no 6 on row 6.
r6c1 can't be 5 as that places 2 in r5c2 and 6 in r5c1, breaking the kropke it's on.
r2c1 can't be 2, because that forces 3o4 into the kropke, and there's no 1 on row 2.
r1c1 can't be 2, because r2c5 also becomes 2, and both of r[12]c6 need to be 1s.
=> Column 1's 2 is in r6c1, creating an X-wing of 2s in r[12]c[25].

r5c1 can't be 5, as both of r5c[25] would become 6s.
=> the [36] pair it creates resolves box 6 with dominoes 5/6 and 2/3.
=> r4c1 can't be 6
=> r3c1 can't be 5, as the 4o3 kropke it induces puts 6 in both of r[45]c2.
=> row 3's 5 is in r3c6

Neither of r[12]c1 can be 4, as the induced 6/5 below breaks the kropke

r1c3 can't be 4, as the 6 forced into r1c6 and the 4 forced into r2c6 mean there's no 1 in column 6.
=> r1c4 can't be 3, creating a [56] pair there, a [14] pair in r[12]c6, and a [23] pair in r[12]c5.
=> boxes 4, 3, and 5 in turn resolve completely
The 1 and 6 forced into box 1 then permit the top two rows to completely resolve.

Proof of uniqueness
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https://sigh.github.io/Interactive-Sudoku-Solver/?q=.Shape%7E6x6.WhiteDot%7ER2C6%7ER2C5.WhiteDot%7ER3C6%7ER4C6.WhiteDot%7ER6C3%7ER6C2.WhiteDot%7ER1C3%7ER1C4.WhiteDot%7ER4C1%7ER5C1.Cage%7E3%7ER3C2%7ER3C3.Cage%7E4%7ER4C3%7ER5C3.Cage%7E5%7ER5C4%7ER6C4.Pair%7EAEC%7E_s5p6%7ER5C6%7ER6C6