Does a Perfect Square Lie Within these 10 Integers?

[myth_kill]

Do there exist 10 distinct integers such that the sum of any 9 of them is a perfect
square.

 

[coolul007]

a = 1/9 (p^2 + q^2 + r^2 + s^2 + t^2 + u^2 + v^2 + w^2 + x^2 - 8 y^2),
b =1/9 (p^2 + q^2 + r^2 + s^2 + t^2 + u^2 + v^2 + w^2 - 8 x^2 + y^2),
c= 1/9 (p^2 + q^2 + r^2 + s^2 + t^2 + u^2 + v^2 - 8 w^2 + x^2 + y^2),
d=1/9 (p^2 + q^2 + r^2 + s^2 + t^2 + u^2 - 8 v^2 + w^2 + x^2 + y^2),
e = 1/9 (p^2 + q^2 + r^2 + s^2 + t^2 - 8 u^2 + v^2 + w^2 + x^2 + y^2),
f = 1/9 (p^2 + q^2 + r^2 + s^2 - 8 t^2 + u^2 + v^2 + w^2 + x^2 + y^2),
g = 1/9 (p^2 + q^2 + r^2 - 8 s^2 + t^2 + u^2 + v^2 + w^2 + x^2 + y^2),
h = 1/9 (p^2 + q^2 - 8 r^2 + s^2 + t^2 + u^2 + v^2 + w^2 + x^2 + y^2),
i = 1/9 (p^2 - 8 q^2 + r^2 + s^2 + t^2 + u^2 + v^2 + w^2 + x^2 + y^2),
j = 1/9 (-8 p^2 + q^2 + r^2 + s^2 + t^2 + u^2 + v^2 + w^2 + x^2 + y^2)

Mathematica did all the donkey work, so now you have to play with perfect squares to resolve each integer.

 

[ramsey2879]

a = 1/9 (p^2 + q^2 + r^2 + s^2 + t^2 + u^2 + v^2 + w^2 + x^2 - 8 y^2),
b =1/9 (p^2 + q^2 + r^2 + s^2 + t^2 + u^2 + v^2 + w^2 - 8 x^2 + y^2),
c= 1/9 (p^2 + q^2 + r^2 + s^2 + t^2 + u^2 + v^2 - 8 w^2 + x^2 + y^2),
d=1/9 (p^2 + q^2 + r^2 + s^2 + t^2 + u^2 - 8 v^2 + w^2 + x^2 + y^2),
e = 1/9 (p^2 + q^2 + r^2 + s^2 + t^2 - 8 u^2 + v^2 + w^2 + x^2 + y^2),
f = 1/9 (p^2 + q^2 + r^2 + s^2 - 8 t^2 + u^2 + v^2 + w^2 + x^2 + y^2),
g = 1/9 (p^2 + q^2 + r^2 - 8 s^2 + t^2 + u^2 + v^2 + w^2 + x^2 + y^2),
h = 1/9 (p^2 + q^2 - 8 r^2 + s^2 + t^2 + u^2 + v^2 + w^2 + x^2 + y^2),
i = 1/9 (p^2 - 8 q^2 + r^2 + s^2 + t^2 + u^2 + v^2 + w^2 + x^2 + y^2),
j = 1/9 (-8 p^2 + q^2 + r^2 + s^2 + t^2 + u^2 + v^2 + w^2 + x^2 + y^2)

Mathematica did all the donkey work, so now you have to play with perfect squares to resolve each integer.

You merely have to make each variable distinct and divisible by 9