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I cannot figure this out:

I have a square in the plane with side length $5$. $A$ and $B$ are points in the square. The coordinates of $A$ and $B$ are always integers.

I want to know how many unique Euclidean distances are possible between $A$ and $B$.

I thought $15$?

Editor note The original ambiguous phrasing of this question and the consequent edits to clarify resulted in conflicting solutions. Please take this into consideration as you vote on the answers.

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    You're not alone, I can't figure out what you're saying either. Distances between what? "Plane of integers?" What does the matrix have to do with this? Please clarify, and include your work on why you think "15".2012-08-10
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    ok the matrix is not the best describtio, but i couldnt come up with a more appropriate tage2012-08-10
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    Do you mean we are looking at the lattice of integer coordinates in a five by five square in the plane? Say, a square with vertices $(0,0),(5,0),(0,5)$ and $(5,5)$? And you want to know what the potential Euclidean distances are between points in this square?2012-08-10
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    yes that is exactly what i mean2012-08-10
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    @rschwieb: This shows why it's a good idea to clearly mark an edit that affects the meaning of a question such that it invalidates existing answers. My answer made sense in response to the original question, which spoke of a $5\times5$ matrix both in the title and in the body; it got two downvotes after your edit.2012-08-11
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    @joriki I'm sorry it turned out that way, but I don't accept the entire blame for this. The original question was incomprehensibly phrased, and I really don't know how you could even venture a solution. At the time I asked for clarification your solution was not up, and by the time I received clarification, your solution was up. I'll do my best to fix the problem, though.2012-08-11
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    @rschwieb: Sorry, I didn't mean to apportion blame for this case; certainly not entire blame :-) Perhaps I sometimes write answers too quickly; this is because I'm annoyed by all the comments that basically solve the question but then when it's confirmed they solve the question noone posts an answer and the question remains unanswered -- I agree that I was perhaps too quick to post in this case. Anyway, thanks for the edit! :-)2012-08-11
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    @joriki No problem: thanks for alerting me to problems I inadvertently caused.2012-08-11

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The problem seems to be solved if you count the number of right triangles with integer length short legs (including length 0 legs) which fit into the square, and then check to see if any of the hypotenuse lengths happen to coincide.

I count 21 such triangles (including degenerate ones), which yield 20 distinct values for the hypotenuse.

The only duplicate distance occured was 5: for example, between (0,0) and (0,5), or else between (0,4) and (3,0).

This is all done with the understanding that we are allowed to pick any integer coordinate point in the square $(0,0), (5,0),(0,5),(5,5)$, so I may be interpreting it differently from other people.

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    Note that for very large grid sizes the duplicates actually dominate: for an $n \times n$ grid there are only about $cn^2 / \sqrt{\log n}$ distances.2012-08-10
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If you count vertical or horizontal distances the values are 0,1,2,3,4,5;

If the horizontal (or vertical) has length 1 and the vertical has length 1, 2, 3, 4, 5 then the distances are $\sqrt{1^2+1^2}=\sqrt{2}, \sqrt{1^2+2^2}=\sqrt{5}, \sqrt{1^2+3^2}=\sqrt{10}, \sqrt{1^2+4^2}=\sqrt{17}, \sqrt{1^2+5^2}=\sqrt{26}$,

If the horizontal (or vertical) has length 2 and the vertical has length 2, 3, 4, 5 then the distances are $\sqrt{2^2+2^2}=\sqrt{8}, \sqrt{2^2+3^2}=\sqrt{13}, \sqrt{2^2+4^2}=\sqrt{20}, \sqrt{2^2+5^2}=\sqrt{29}$,

If the horizontal (or vertical) has length 3 and the vertical has length 3, 4, 5 then the distances are $\sqrt{3^2+3^2}=\sqrt{18}, \sqrt{3^2+4^2}=\sqrt{25}=5, \sqrt{3^2+5^2}=\sqrt{34}$.

If the horizontal (or vertical) has length 4 and the vertical has length 4, 5 then the distances are $\sqrt{4^2+4^2}=\sqrt{32}, \sqrt{4^2+5^2}=\sqrt{41}$.

If the horizontal (or vertical) has length 5 and the vertical has length 5 then the distances are $\sqrt{5^2+5^2}=\sqrt{50}$.

So there are 20 distinct values for the distances between two points.

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To the downvoters: This answer was in response to the original question, which asked about a "$5$ by $5$ matrix". The answer was invalidated by an edit that affected the meaning of the question and wasn't marked as such.


I count $15$, too. What makes you think that's wrong?

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    well nothing but i wanted a formal proof but i see it know.2012-08-10
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    @jorrebor: Please take more care phrasing your question next time. Your question states twice that you want to know the number of unique distances, and nowhere that you wanted a formal proof.2012-08-10
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    yes you're right2012-08-10