Does a Perfect Square Lie Within these 10 Integers?

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Do there exist 10 distinct integers such that the sum of any 9 of them is a perfect
square.
 
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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.
 
coolul007 said:
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
 

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