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?
 A: This answer is only for the case $N=4$. I fail to see how this method could help resolve the cases for larger $N$.
Consider the equations
$$a + b = u^2, c + d = v^2$$
$$a + c = w^2, b + d = x^2$$
$$a + d = y^2, b + c = z^2$$
$$a + b + c + d = k^2$$
Now the first three pairs of equations give
$$u^2 + v^2 = w^2 + x^2 = y^2 + z^2 = k^2$$
so a reasonable approach would be to investigate Pythagorean triples.
Not all Pythagorean triples generate positive integers solutions, but a solution could be generated if the pairs $(u,v), (w,x), (y,z)$ are "reasonably close". Consider
$$165^2+280^2 = 195^2+260^2 = 253^2+204^2 = 325^2$$
Take note of the parity as well (or multiply it by $2$ so you don't need to worry about it)
Solving the equations give
$$a = 11817, b = 15408, c = 26208, d = 52192$$
and more solutions can be generated this way.
A: The solution given by "@ player3236" has $k=325$.
The integer $325$ can be written as sum of two squares
in three way's & is shown below:
$k= x_1^2+x_2^2=x_3^2+x_4^2=x_5^2+x_6^2$
($x_1,x_2,x_3,x_4,x_5,x_6$)=$(15,10,17,6,18,1)$
$325= 15^2+10^2=17^2+6^2=18^2+1^2$
Above can help in arriving at a parametric solution.
Also mathematician "Seiji Tomita has given parametric
solution in which $(a,b,c,d)$ taken two at a time is a
square & the link is given below.
http://www.maroon.dti.ne.jp/fermat/dioph115e.html
And click on (dioph115e)
A: 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:
$(a,b,c,d)=(39169,-23544,24840,65160)$
