Is it possible to find three number $a, b, c$ such that sum of any two gives you a square number? For example $6+3=9=3^{2}$. And what is the formula to find these numbers?
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It is quite easy to find a formula giving you all possible triples: $$ a={x^2-y^2+z^2\over2},\quad b={x^2+y^2-z^2\over2},\quad c={-x^2+y^2+z^2\over2}. $$ The sum of the three squares must be even, so $x$, $y$ and $z$ must be all even, or one even and two odd. In addition, for $a$, $b$ and $c$ to be positive $x^2$, $y^2$ and $z^2$ must respect the triangular inequality: each of them must be less than the sum of the other two.
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Yes, an example is $(10/54/90)$
The following PARI/GP program finds some triples :
? for(a=1,60,for(b=a+1,60,for(c=b+1,60,if(issquare(a+b)*issquare(a+c)*issquare(b
+c)==1,print(a," ",b," ",c)))))
2 34 47
4 21 60
5 20 44
6 19 30
16 33 48
?
There is also a constructive way :
Choose the squares $k^2,l^2,m^2$, such that $0<k<l<m$ and $k^2+l^2>m^2$
Then, the triple $$(\frac{k^2+l^2-m^2}{2}/\frac{k^2-l^2+m^2}{2}/\frac{-k^2+l^2+m^2}{2})$$ does the job.
