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I'm solving one hard problem in my homework textbook (it is from the list of hardest problems in the end of book with stars). I reduced it to very simple question which I can prove by two, but very complicated and long ways (using Heron formula and some long algebra operations).

It must be easy and simple (I hope) solution to this question, which i can't see.

Question is: We have two parallel lines $(l_1,l_2)$ and the distance between this lines $|DE|=n$ an integer(see pic.), $n \in \mathbb{N}$. Let points $A$,$B \in l_1$ and $|AB|=|DE|=n$. Let point $C \in l_2$ and $|AC|=k,|BC|=m$. Prove that there are no exist such point $C$ that $k$ and $m$ both integer. (If $k,n \in \mathbb{N}$, then $m \notin \mathbb{N}$ or if $m,n \in \mathbb{N}$, then $k \notin \mathbb{N}$).

I can prove it (like i said before it is very long analysis of equation which we can obtain using formulas for area), I'm looking for simple and short solution. Thanks.

enter image description here

In my proof I use equation

$$ 4n^4=(n+k+m)(k+m -n)(n+k-m)(n-(k-m)), \ \text{if } x=k+m, y=k-m \Rightarrow $$

$$ 5n^4-(x^2+y^2) n^2 +x^2 y^2=0. $$

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    More a suggestion than an answer, since I haven't fleshed this out all the way yet, but: if you let $p$ be the distance between $B$ and the projection of $C$ down onto $\ell_1$, then $p$ has to be an integer and in fact $(n, p, m)$ and $(n, n+p, k)$ are both primitive Pythagorean triples, implying that $n$ must be even; maybe the problem can be attacked from that angle?2012-09-04
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    I consider that. It is easy to prove that if p integer, then such Pythagorean triples doesn't exist. But p not necessarily integer. At least I don't know how to prove that p has to be integer. I think it just particular case.2012-09-04
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    Actually, it's easy to prove that $p$ is an integer: $k^2=n^2+(n+p)^2 = n^2+p^2+2np+n^2 = m^2+n^2+2np$, and since $m$, $n$ and $k$ are integers then $2np$ is also an integer, so $p$ is rational; but since $p$ is rational and $p^2=m^2-n^2$ is an integer, then $p$ is an integer.2012-09-04
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    Oh! Thank you! I think this is what I need. I knew that there is some easy nice way, better then this stupied equation.2012-09-04
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    I didn't even thought about this wonderful step *(p is rational; but since p is rational and $p^2=m^2−n^2$ is an integer, then p is an integer)*. I was sure that p can be rational.2012-09-04
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    You could break up the problem into 2 right $\Delta$ 's by drawing a line from $A$ perpendicular to $l_2$ and a line from $C$ perpendicular to $l_1$. From there it will be a simple matter of the trigonometry of right $\Delta$ 's2014-02-07
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    @StevenStadnicki I am confused by your assumption. He is trying to prove the there does not exist any triplet $(n,m,k)$ such that each is an integer, and you are assuming that each is an integer in order to show that p is an integer. Am I misinterpreting something?2014-03-29
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    @Daniel I think you are; I'm suggesting a proof by _contradiction_: assume that $(n,m,k)$ are all integers. Then $p$ must also be an integer, and you have the triples I mention; this may be enough to generate the requisite contradiction.2014-03-29
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    @StevenStadnicki Ah, I see. Thanks for the clarification. Another similar idea, perhaps not leading anywhere, would be the following: assume each is an integer, and without loss of generality assume $k \geqslant m$, then $\exists$ a $\alpha \in \mathbb{N}$ s.t. $k = m + \alpha$. From here, work with the geometry to show that $\alpha$ cannot be an integer.2014-03-29

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