Geometry - Rectangle ABCD with inside point E. Find the least possible value for sum of interger distances from E to 4 vertices. 
The point E lies within the rectangle ABCD.
If the distances from the vertices to E are all distinct integers, what is the least possible value of AE + BE + CE + DE?
 A: We want
$$\left\{
\begin{align*}
a^2&+c^2&=p^2& \hspace{4em} (1)\\
a^2&+d^2&=q^2 &\hspace{4em} (2)\\
b^2&+c^2&=r^2 &\hspace{4em} (3)\\
b^2&+d^2&=s^2 &\hspace{4em} (4)
\end{align*}
\right.$$
where $a+b$ and $c+d$ are the side lengths of the rectangle, and $p,q,r,s$ are the distances to the vertices.  We want to find a way such that $p,q,r,s$ are distinct integers (i.e. let $\gcd(p,q,r,s)=1$).
\begin{align*}
(1)-(2): && c^2-d^2&=p^2-q^2 &\hspace{4em} (5) \\
(3)-(4): && c^2-d^2&=r^2-s^2 &\hspace{4em} (6) \\
(5)-(6): && 0 & = p^2-q^2-r^2+s^2 \\
&&r^2-s^2 &=p^2-q^2 & \hspace{4em}(7)
\end{align*}
A solution (and the solution containing the smallest numbers) for the diophantine equation $(7)$ is
$$(p,q,r,s)=(8,4,7,1).$$
Therefore, the answer to your question is the sum of $p,q,r,s$, or $20$.
More info:  https://oeis.org/A118882
A: Hint: 
Recognize the relationship
$$AE^2+CE^2=BE^2+DE^2$$
which would well reduce the guess work.
A: Suppose we have such rectangle.  Place it on a coordinate system so that the $E$ is on $(0,0)$ and $A = (-a, b); B= (-a,-c), C=(d,b); D=(d,-c)$ and we need for $\sqrt{a^2 + b^2}=m, \sqrt{a^2+c^2}=n, \sqrt{d^2 + b^2}=r, \sqrt{d^2+c^2}=s$ are all integers.
We must have $m^2 +s^2 = n^2 + r^2$ and that simply requires finding the smallest integer that is a sum of two distinct perfect squares in two possible ways that will have the smallest sum.
$m^2 + s^2 = n^2 + r^2 \implies m^2-n^2 = r^2 - s^2 \implies (m+n)(m-n)=(r+s)(r-s)$.
Now $m+n, m-n$ must be the same parity and and we could do $r+s=5, r-s=3$ and $m+n=15, m-n=1$ so $m=8;n=7;r=4;s=1$.  This would be the smallest set of four distinct  values for $m-n,r-s,r+s,m+n$.  And so $s+r+n+m=20$.
And $8^2 + 1^2 = 7^2 + 4^2=65$ is a probably this smallest such integer.
(I'm not sure how to verify this other than trial and error.  Oddly enough taking even values $m-n=2, r-s=4,r+s=6,m+n=12$ so $r=5;s=1;n=5;m=7$ gives us a smaller integer $7^2+1=5^2 + 5^2=50$ and as smaller sum of $18$ but the terms are not distinct.)
So we need $a^2 + b^2= 64; a^2+c^2=49; d^2+b^2=16; d^2 + c^2 =1$.
$d^2=1-c^2$ and so $a^2 + b^2 = 64; a^2+c^2=49; b^2-c^2=15$
$b^2 = 15+c^2$ and so $a^2+c^2=49$. So we need $0< c^2 <1$ the$1> d^2=1-c^2>0$, $b^2 = 15+c^2> 15$ and $a^2 = 49-c^2> 48$ are possible rectangles.
