Biggest set such that sum of any pair is perfect square What is the biggest set of positive integers such that the sum of any pair of them is a perfect square? (Or can we construct an infinite such set?)
One such set of size $3$ is $\{6,19,30\}$, which give sums $25,36,49$, but I can't find any larger yet.

It's easy to do. The formula below. It is only necessary to take $a,b,c$.
You can even for 4 terms to write.
Do there exist four distinct integers such that the sum of any two of them is a perfect square?  This is equivalent to solving the following system of equations:  $$\left\{\begin{aligned}& b+a=x^2 \\&b+c=y^2\\&b+f=z^2\\&a+c=e^2\\&a+f=j^2\\&c+f=p^2\end{aligned}\right.$$  Let: $F,T,R,D$ - any asked us integers.  For ease of calculation, let's make a replacement.  $$q=(8F^2+4FT-T^2)R^2+2(T+2F)RD-D^2$$  $$k=(8F^2+8FT+2T^2)R^2+2(T+2F)RD$$  $$s=-T^2R^2+2(T+2F)RD-D^2$$  $$t=(8F^2+12TF+3T^2)R^2+2(T+2F)DR-D^2$$  Then the solutions are of the form:  $$x=s^2+k^2-t^2+2(t-k-s)q$$  $$y=t^2+k^2-s^2+2ks-2tk$$  $$z=s^2+k^2-t^2$$  $$e=t^2+k^2+s^2-2kt-2ts$$  $$j=t^2+s^2-k^2+2ks-2ts$$  $$p=3s^2+3k^2+3t^2-6kt-6st+8ks+2(t-k-s)q$$  $$b=\frac{x^2+y^2-e^2}{2}$$  $$a=\frac{e^2+x^2-y^2}{2}$$  $$c=\frac{e^2+y^2-x^2}{2}$$  $$f=\frac{2z^2+e^2-x^2-y^2}{2}$$
