<s>[INST] Can this problem have solutions for $N>3$:Find three $(N=3)$ positive integers a, b and c such that: $a+b+c=k^2$, $a+b=t^2$, $b+c=m^2$ and $a+c=n^2$ Can this question have solutions for N>3:
Find three ($N =3$) positive integers a, b and c such that: $a+b+c=k^2$, $a+b=x^2$, $b+c=m^2$ and $a+c=n^2$; $x, m, n∈\mathbb   N$
Solution:
Let $a+b+c=(x+1)^2=x^2+2x+1$
If $a+b=x^2$ then third number $c=2x+1$.
If $b+c=(x-1)^2=x^2-2x+1=m^2$ then :
$a=4x$, so $b=x^2-4x$
Now due to statement we must have $a+c=6x+1=n^2$
$n^2=6x+1$ can have infinite solutions such as:
$(x, n^2)=(20, 121=11^2), (60, 361=19^2), (140, 841=29^2)\cdot\cdot\cdot$
Which give:
$(a, b, c, k)=(80, 320, 41, 21), (240, 3360, 121, 61), (560, 19040, 281, 141),\cdot\cdot\cdot$
I tried to solve this problem for $N=4$, four numbers but no success. Now I have two questions:
1-Does this problem have solutions for N>3?
2-Any idea for better method?
 [/INST] We have:
$a + b = u^2$, $c + d = v^2$
$a + c = w^2, b + d = x^2$
$a + d = y^2, b + c = z^2$
$a + b + c + d = k^2$
Let:
$u=m^2-n^2$ & $v =2mn$
$w=p^2-q^2$ & $x=2pq$
$y=r^2-s^2$ & $z=2rs$
We apply the condition:
$m^2+n^2=p^2+q^2=r^2+s^2=k^2$
Hence, $u^2+v^2=w^2+x^2=y^2+z^2=k^2$
Since:
$2a=u^2+w^2-z^2$
$2b=x^2+z^2-v^2$
$2c=v^2+w^2-y^2$
$2d=v^2+x^2-z^2$
Substituting value of $(u,v,w,x,y,z)$ from above we get:
$a=1/2(m^4+n^4+p^4+q^4)-(m^2n^2+p^2q^2+2r^2s^2)$
$b=2(p^2q^2+r^2s^2-m^2n^2)$
$c=1/2(p^4+q^4-r^4-s^4)+(2m^2n^2-p^2q^2+r^2s^2)$
$d=2(m^2n^2+p^2q^2-r^2s^2)$
For: $(m,n,p,q,r,s,k)=(15,10,17,6,18,1,325)$ we get:
$(a,b,c,d)=(39169,-23544,24840,65160)$
</s>