$\Large\lfloor \sqrt{x} \rfloor + \lfloor \sqrt{y} \rfloor + \lfloor \sqrt{z} \rfloor = 140$

Find the number of non-negative integral solutions to the equation above.

**
Note:
**
$\lfloor k \rfloor$
represents the greatest integer less than or equal to
$k$
.

You can use calculators to calculate the final sum.

The answer is 3782205855.

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The number of non-negative integer solutions to $\lfloor \sqrt{x} \rfloor + \lfloor \sqrt{y} \rfloor + \lfloor \sqrt{z} \rfloor = 140$ is equivalent to finding the coefficient of the term $t^{140}$ in the expansion,

$(1+t^{\lfloor \sqrt{1} \rfloor} + t^{\lfloor \sqrt{2} \rfloor} +\ldots )(1+t^{\lfloor \sqrt{1} \rfloor} + t^{\lfloor \sqrt{2} \rfloor} + \ldots )(1+t^{\lfloor \sqrt{1} \rfloor} + t^{\lfloor \sqrt{2} \rfloor} + \ldots )$

$=(1+t + t+ t+ t^{2} +t^{2} + t^{2} + t^{2} + t^{2} + \ldots)^{3}$ $=(1+3t + 5t^{2} +7t^{3} + \ldots )^{3}$ $=\left( \dfrac{1+t}{(1-t)^{2}} \right) ^3 \text{(Sum of terms in infinite A.G.P)}$ $=(1+t)^{3} (1-t)^{-6}$ $=(1+3t+3t^{2} + t^{3})(1+ \binom{6}{1} t + \binom{7}{2} t^{2} + \ldots + \binom{5+r}{r} + \ldots )$

So, the coefficient of $t^{140}$ in the above expansion is

$\large{\binom{145}{140} + 3 \binom{144}{139} + 3 \binom{143}{138} + \binom{142}{137}}=\boxed{3782205855}$ .