Whats wrong/illegal in this position?

What is a wrong/illegal thing in this diagram? Hint: There are multiple things.

Two black bishops for white, two white bishops for black. Pawn on d6 for black without any pieces being taken.

More interesting: What is the least number of units that have to be removed to make it a legal position?

two - I think. Remove one of white's bishops then one of blacks.  White's bishop could (theoretically) have been taken on d6

No, d6 is a dark square; it couldn't have captured White's light bishop.

Three then. Remove both White bishops and one Black bishop. Other combinations are also valid.

good point, yeah 3 then

If "Pd2 were on d4", only 2 units would need to be removed to make the diagram legal. You are clever if you can see how and why.

how? why? I can't see after 4 minutes looking

Remove Pawn d7 and a black squared Bishop. In the proof game:

1. White first plays pawn to d4

2. Black plays e7-pawn to e4

3. Black captures white squared bishop: e4xBd3

4. Pawn d3 promotes on d1 to become the 2nd white squared bishop

Nope to that. White has 2 dark squared bishops. 3 units need to be removed in any case.

no, I think arisktotle's idea does work.

You are both right! My idea won't work in the diagram, only after pawn d2 is moved to d4 (post #8)

...If wPd2 is moved to d4, white still has 2 dark squared bishops on b4 and e3, one of which must be promoted. At least 3 units need to be removed.

Queen and bishop are both hanging for white. No time like the present for white to resign.

See my post #8 and #10. I removed Pd7 plus one white black squared bishop, so only one white black squared bishop remains.

By the way, congrats on your nice 'bishop Belfort'. I posted a comment on it.

Ah, I misread. I read it as "black bishop" (incidentally why I use light/dark squared bishops.)

I have some bad habits. Your way is better

how did e3 bishop get there anyway??  from what starting square?

It arrived by moonlander. For the rest of the story, see post #1.