Squares, Floors and Fractions? Pfft!

By definition, the floor function is an integer, so in this case 38 is the sum of 4 squares. The only squares for which this holds are 16 + 9 + 9 + 4 = 38. Considering the first (and largest) square, we know $\lfloor \frac {x_{1}+x_{2}} {3} \rfloor = 4$ in order for the square itself to equal 16. Therefore, due to the nature of the floor function,

$4 \leq \frac {x_{1}+x_{2}} {3} < 5$

so

$x_{1} + x_{2}$ is 12, 13 or 14

Applying the same process for the other three squares, we obtain the following possibilities.

$x_{2} + x_{3}$ is 9, 10 or 11

$x_{3} + x_{4}$ is 9, 10 or 11

$x_{4} + x_{5}$ is 6, 7 or 8

We know $x_{2} > x_{3} > x_{4}$ , so because they are integers, $x_{2}$ must be at least 2 bigger than $x_{4}$ . Therefore $x_{2} + x_{3}$ must be at least two bigger than $x_{3} + x_{4}$ . The only possibility which fits this is $x_{2} + x_{3} = 11$ and $x_{3} + x_{4} = 9$

We know that $x_{4} > x_{5}$ , so in order for $x_{4} + x_{5}$ to be 6, 7 or 8, it is necessary that $x_{4} \geq 4$

We also know that $x_{3} > x_{4}$ and that $x_{3} + x_{4} = 9$ from before. Because $x_{4} \geq 4$ , the only possibility is that $x_{3} = 5$ and thus $x_{4} = 4$

We know that $x_{2} + x_{3} =11$ , so it follows that $x_{2} =6$

This means $x_{1}$ and $x_{5}$ have two possibilities each. $x_{1}$ could equal 7 or 8, and $x_{5}$ could equal 2 or 3. We now have all the different possible values for the numbers. To summarise the four different cases, the values for $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ are:

7, 6, 5, 4, 2

7, 6, 5, 4, 3

8, 6, 5, 4, 2

8, 6, 5, 4, 3

The question wants the sum of the numbers in each of the possible solutions, which is

$(7+6+5+4+2)+$

$(7+6+5+4+3)+$

$(8+6+5+4+2)+$

$(8+6+5+4+3)$

$= 24+25+25+26=100$

Let $\lfloor \frac{x_{i}+x_{i+1}}{3} \rfloor=a,b,c,d$ when $i=1,2,3,4$ respectively. Now we have :

• $a \geqslant b \geqslant c \geqslant d \text {(at most one inequality hold)}$

• $a^{2}+b^{2}+c^{2}+d^{2}=38$

So the only values of $a,b,c,d$ is $4,3,3,2$ respectively.

Since $b=c$ we can claim that $(x_{4},x_{3},x_{2})=(x_{4},x_{4}+1,x_{4}+2)=(4,5,6)$

From $a=4$ and $d=2$ we get $x_{1}=(7 or 8)$ and $x_{5}=(2 or 3)$ .

the solutions are : $(2,4,5,6,7) , (2,4,5,6,8)$ $(3,4,5,6,7) , (3,4,5,6,8)$ Answer: $(4+5+6) \times 4+14+16=\boxed {100}$

---

note: writing a solution by a phone it might be not the best idea :P

Inorder to find the no.of ways in which a no. Can be expressed as a sum of 4 squares one may use Euler's 4 square identity. Their are many more methods to do the same.