Find a triangle with integer sides and medians.

**
There exists only one generic solution! **

(In reply to

contradictions? by Charlie)

Apparently, there is more than one generic solution.

Ch,

I have checked the compliance of your triplets with the formula:

mc=.5*sqrt(2*a^2+2*b^2-c^2) and found the following:

(68, 85, 87**)=>(136, 170, 174**) with medians 158, 131 and 127

(127, 131, 158) **a wrong triplet** (medians of the previous)

(113, 243, 290)=>(**226,486,580**) with medians 523,367 and 244

(145, 207, 328)=>(**290,414,656**) with medians 529,463, and 142

**(**327, 386, 409**)=>(654,772,818**) with medians 725,632, & 587.

The original question (from a Russian contest) was neither solved

nor altered by me.** The claim of uniqueness was there.**

I believe that the only way to solve it is checking all triplets, say

below 1000, s.t. a<b<c<(a+b), rotating the values of the triplet,

applying the formula up to 3 times, and saving the "all integers" results.

**Once again both of us learn not to be misguided by some web authority.**