MHB Lengths of the sides of quadrilateral

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In a quadrilateral with positive integer side lengths, each side must divide the sum of the other three sides. This condition leads to a relationship among the sides that implies at least two sides must be equal. The proof relies on the properties of divisibility and the constraints imposed by the integer nature of the side lengths. By analyzing the implications of the divisibility condition, it can be shown that the equality of two sides is necessary. Therefore, it is proven that in such a quadrilateral, at least two sides must have the same length.
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The lengths of the sides of a quadrilateral are positive integers. The length of each side divides the sum of the other three lengths. Prove that two of the sides have the same length.
 
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If the quadrilateral has sides $a,b,c,d$ and perimeter $s = a+b+c+d$ then $a < b+c+d$. Also, $a$ divides $b+c+d$, so $b+c+d$ is at least twice $a$. But $b+c+d = s-a$, so $a$ also divides $s$, and $s$ must be at least $3$ times $a$, say $s=ka$ with $k\geqslant3$. Similarly, $s = lb$, $s=mc$ and $s=nd$, with all of $l,m,n$ greater than or equal to $3$.

Then $a = \dfrac1ks$, $b = \dfrac1ls$, $c = \dfrac1ms$, $d = \dfrac1ns$. Therefore $s = a+b+c+d = \dfrac1ks + \dfrac1ls + \dfrac1ms + \dfrac1ns$ and so $$\frac1k + \frac1l + \frac1m + \frac1n = 1.$$ But under the given conditions, if $a,b,c,d$ are all different then so are $k,l,m,n$, and the largest possible value for $\dfrac1k + \dfrac1l + \dfrac1m + \dfrac1n$ is $\dfrac13 +\dfrac14 + \dfrac15 + \dfrac16 = \dfrac{57}{60}$, which is less than $1$. So those conditions cannot be satisfied and therefore at least two of the four sides must have the same length.
 

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