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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.

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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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1  
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? –  Steven Stadnicki Sep 4 '12 at 22:14
    
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. –  Mike Sep 4 '12 at 22:22
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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. –  Steven Stadnicki Sep 4 '12 at 22:26
    
Oh! Thank you! I think this is what I need. I knew that there is some easy nice way, better then this stupied equation. –  Mike Sep 4 '12 at 22:32
    
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. –  Mike Sep 4 '12 at 22:40

2 Answers 2

Let $\phi\equiv$ angle subtended by $ACB$

$\leadsto k^2+m^2=n^2-2km\cos\phi$ (1)

area $\equiv\frac{(base)(foot)}{2}=\frac{n^2}{2}$

$\therefore\frac{n^2}{2}=\frac{1}{4}\sqrt{(k^2+m^2+n^2)^2-2(k^4+m^4+n^4)}$

This is Heron's formula. It involves nothing very obscure so far. Square to clear the root. Multiply by 16. Your LHS obtains.

$4n^4=(k^2+m^2+n^2)^2-2(k^4+m^4+n^4)$ (2)

Add $n^2$ to (1). Substitute RHS for LHS in (2).

$4n^4=(2n^2-2km\cos\phi)^2-2(k^4+m^4+n^4)$

$4n^4=4n^4-8n^2km\cos\phi+4k^2m^2\cos^2\phi-2(k^4+m^4+n^4)$

Rearranging,

$8n^2km\cos\phi-4k^2m^2\cos^2\phi+2(k^4+m^4+n^4)=0$

and

$n^4+4(km\cos\phi)n^2+k^4+m^4-2k^2m^2\cos^2\phi=0$ (3)

which is biquadratic in $n^2.$

(3) $\leadsto n^2=\frac{-km\cos\phi}{2}+\sqrt{6k^2m^2\cos^2\phi-(k^4+m^4)}$

where $n^2$ is a perfect square. By symmetry, let $k>=m$ without loss of generality. If $k=m$ then

$n^2=\frac{-k^2\cos\phi}{2}+\sqrt{k^4(6\cos^2\phi-2)}$, $n^2=k^2(\sqrt{6\cos^2\phi-2}-\frac{\cos\phi}{2})$

By hypothesis both $n$ and $k$ are elements of $\mathbb{N}$, but their ratio is $\sqrt{\sqrt{6\cos^2\phi-2}-\frac{\cos\phi}{2}}$. So maybe the hypothesis doesn't hold for this case.

This would be the simplest case. However by hypothesis, $k=m+p:p\in\mathbb{N}$ so we should be able to show the ratio $n:k$ is not correct in any case.

Maybe that's not very short, but it's simple.

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A beginning:

When this can be done for integer $n$ such that $k$ and $m$ are integers, then it can be done with $n$ even. Therefore I choose $$A=(-n,0),\quad B=(n,0),\quad C=(x,2n)$$ with $x\geq 0$. Then $$k=\sqrt{4n^2 +(x+n)^2},\quad m=\sqrt{4n^2 +(x-n)^2}\ .\tag{1}$$ Squaring and subtracting gives $$x={k^2-m^2\over 4n}\ ,$$ which proves that $x$ has to be rational. But we can say more: From $(1)$ it also follows that $$x=-n+\sqrt{k^2-4n^2},\qquad x=n+\sqrt{m^2-4n^2}\ .$$ The two surds can only be rational if $k^2-4n^2$ is a square $k'^2$, and similarly $m^2-4n^2$ is a square $m'^2$. As a consequence $x$ will be an integer.

Therefore the problem boils down to the following: We have to prove that the diophantine system $$k^2=k'^2+4n^2,\quad m^2=m'^2+4n^2,\quad 2n=k'-m'$$ has no nontrivial solutions. One would have to look at the theory of pythagorean triples $\ldots$

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