How many moves does a knight need to get somewhere?

On an 8x8 it should take 6 max

Does that not remind you of somebody else?

I ROUNDHOUSE KICK YOU

But to answer you question, no, not really

Get somewhere does mean go somewhere, so it can't be no moves cos then it wouldn't get anywhere at all, which is a subset of everywhere. It has to be one.

I mean... that suggests that objects that don't move don't exist in a space, that you can only exist if you move to a location.

It's all rather silly. No one actually cares.

Disclaimer... I'm drunk right now

His seems efficient. The wave patterns I made are no thicker than 4, and he has two different statements that use division and a floor function meaning each statement could yield two different answers, and 2x2=4.

There are a few special cases he takes care of (when the destination is very close).

So it seems very efficient. There may be some tricks to make it even more efficient, but I'm not experienced with anything like that, so I wont even try.

As the originator of this thread, I can verify that the question in its title was intended to get people's attention, and that the actual intent of the question was spelled out in the initial post clearly enough for many people to understand it. One reader in particular was so unnerved by the idea of an infinite chessboard that he declared the question meaningless. For his benefit, I provided an alternate version of the question in which the board is not infinite, but whose size is so large that a knight near its center cannot have its shortest path to a specified target square influenced by the edge of the board, and noted that it's easy to prove that there is such a size for any given target square.

Later a few armchair philosophers weighed in with comments addressing only the question in the title, and ignoring its intended meaning as explained in the initial post. I don't have any objection to those comments, but I am curious about whether is has occurred to those making them to attempt a positive contribution to the actual intent of the question as explained in the initial post. At this point I don't believe that's likely to happen, but maybe one of them will surprise me.

i have a very simple solution. we give the knight self awareness and force it to move to the square at gunpoint. it should get there pretty fast.

just count how many jumps it makes

Good question...

Nc2-d4-b3-d2-c4 is 4 moves, but that isn't the shortest path. Nc2-e3-c4 is shorter.

Chess.com forum's largest two groups:

1) Kids
2) Retirees slowly slipping into dementia

Incorrect- I will shortly provide the explanation.

The graphs provided by llama elegantly express the shortest Knight moves - on an infinite plane of 360 degrees. 4 quadrants - mirror images. The  expanding symmetry is easily viewed. A formula easily written than can be applied for any starting point. 
However- 

In practical matters- with a chess board that has boundaries which limits the options of possible moves- the chart  is at best 50% correct !

A formula that works on the 360 % chart does not apply to a single quadrant (as in a chess board.)

It is soon realized that the resulting depiction provided by the graphs shortest Knight moves is not at all what the true story is on a chess board.  It seemed the OP was looking to solve a real life math problem to be applied to chess on a practical matter.

An elegant chart depicting the shortest Knight moves provide by llama

Divide the chart on x/y axis into four quadrants. All symmetrical. The resulting picture of each quadrant is not a true representation of the shortest Knight moves of a bordered chess arena. (By the nature of the Knight move). If a Queen - the repeating logarithm works, but not for the Knight.

so you are saying

4 quadrants= equal

4 quadrants alone= different from shown

so, diagram=false?

I couldn't decide what you meant by this, but it appears to me that the pattern is perfectly symmetrical. In checking the symmetry, you have to ignore the leftmost column of pixels, since the corresponding rightmost column is off the edge of the graph. Likewise the bottom row of pixels matches a top row that is off the edge of the graph.

As I suggested, a positive contribution is not something I will hold my breath while waiting for. I'm happy about the very positive contributions already made by some, particularly those of llama47.

Actually, the OP had no interest at all in applying the problem solution to chess as a practical matter. The OP was looking for an efficient way to solve an interesting mathematical problem that had been posed but not solved by someone else in another forum.

The mention of an infinite chessboard in the initial post might have been a clue that the problem was not about real chessboards and practical chess. (And please remember that the same problem with the same solution can be framed in a way that does not require any mention of infinity.)