${{(x+1)}^{2}}+{{(x+2)}^{2}}+...+{{(x+9)}^{2}}={{y}^{2}}\Leftrightarrow 9{{x}^{2}}+9x+15\cdot 19={{y}^{2}}\Rightarrow y=3z\Rightarrow 3{{x}^{2}}+3x+5\cdot 19=3{{z}^{2}}$ contradiction

@above,how did you get $y=3z\Rightarrow 3{{x}^{2}}+3x+5\cdot 19=3{{z}^{2}}?$

Just I replaced y=3z

Let $x+5=n$, we have $(n-4)^2+(n-3)^2+\cdots+(n+4)^2=9n^2+60=y^2$. (Motivation behind this is to cancel terms) This can never be a square, since if $3|y^2$, then $9|y^2$, but $9\nmid 9n^2+60$, thus, there are no solutions.

parmenides51 wrote: Prove that there are no positive integers $x, y$ such that: $(x + 1)^2 + (x + 2)^2 +...+ (x + 9)^2 = y^2$ \[\frac{(x+9)(x+10)(2x+19)}{6}-\frac{x(x+1)(2x+1)}{6}=\frac{54 x^2 + 540 x + 1710}{6}\]\[9 x^2+ 90 x + 285 =y^2\implies 9 x^2+ 90 x + 225 -y^2=-60\]\[(3x+15+y)(3x+15-y)=60\]Use SFFT to prove there is no solution for $\mathbb N$

mod9 is enough