140 Followers Problem

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

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

My solution is more pedestrian than Surya's.

We start with the observation that there are $2p+1$ integers $x$ with $\lfloor{\sqrt{x}}\rfloor=p$ , namely, $p^2\leq{x}\leq{p^2+2p}=(p+1)^2-1$ .

The number of integer pairs $(x,y)$ with $\lfloor{\sqrt{x}}\rfloor+\lfloor{\sqrt{y}}\rfloor=q$ is $\sum_{p=0}^{q}(2p+1)(2q-2p+1)=\frac{1}{3}(q+1)(2q^2+4q+3)$ .

Finally, the number of integer triples $(x,y,z)$ with $\lfloor{\sqrt{x}}\rfloor+\lfloor{\sqrt{y}}\rfloor+\lfloor{\sqrt{z}}\rfloor$ $=140$ is $\sum_{q=0}^{140}\frac{1}{3}(q+1)(2q^2+4q+3)(281-2q)$ $=\boxed{3782205855}$

We have that in order for $\left\lfloor \sqrt { x } \right\rfloor =n$ the following must be satisfied: $n\le \sqrt { x } <n+1$ Squaring both sides we have that: ${ n }^{ 2 }\le x<{ n }^{ 2 }+2n+1$ There are $2n+1$ values that $x$ can take on to statisfy this condition. We can now take a sum to find the total combinations for $x$ , $y$ and $z$ : $\sum _{ X=0 }^{ 140 }{ \sum _{ Y=0 }^{ 140-X }{ (2X+1)(2Y+1)(2(140-X-Y)+1) } }$ Here $X$ denotes the value of $\left\lfloor \sqrt { x } \right\rfloor$ for ${ X }^{ 2 }\le x<{ X }^{ 2 }+2X+1$ , and so does $Y$ . Evaluating the sum we have that the number of combinations of $x$ , $y$ and $z$ is: $\sum _{ X=0 }^{ 140 }{ \sum _{ Y=0 }^{ 140-X }{ (2X+1)(2Y+1)(2(140-X-Y)+1) } }=3782205855$