More Difficult System of Equations

Let $\{x\}$ be the fractional part of $x$ ; that is, $\{x\} = x - \lfloor x \rfloor$ . Since $a,b,c,d$ are integers, $\{a\} = \{b\} = \{c\} = \{d\} = 0$ . Also, $\{x+y\} \equiv \{x\} + \{y\} \pmod{1}$ , which means the fractional part function is additive modulo $1$ . Thus, taking the fractional part of both sides from equation 1,

$\{a+b+c+\sqrt{d}\} \equiv \{a\} + \{b\} + \{c\} + \{\sqrt{d}\} \equiv 0 \pmod{1}$

$\{\sqrt{d}\} \equiv 0 \pmod{1}$

Since $0 \le \{\sqrt{d}\} < 1$ , we obtain $\{\sqrt{d}\} = 0$ .

By the same reasoning, we obtain $\{\sqrt{c}\} \approx 0.208973, \{\sqrt{b}\} \approx 0.057628, \{\sqrt{a}\} \approx 0.207436$ .

Now, since $a,b,c$ are positive integers not greater than $1024$ , we can do a brute force search:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11

# Python 3.4
for num in range(1, 1025):
    sqrt = num**.5
    rem = sqrt - int(sqrt)
    rem = str(rem)[2:8] # from 0.abcdefghijk, we take abcdef
    if rem in ["208973", "057628", "207436"]: print(num, rem)

# prints:
# 586 207436
# 679 057628
# 974 208973

Thus $a = 586, b = 679, c = 974$ , and by plugging to the first equation, we obtain $\sqrt{d} = 2$ or $d = 4$ . Thus the solution is $(a,b,c,d) = (586,679,974,4)$ , which can be verified to be indeed a solution, and thus $1024^3a+1024^2b+1024c+d = \boxed{629925689348}$ .

As stated, I do not have a solution but I will put the value of a,b,c,d here so that people may check that the answer is correct and I do not forget them:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12

In[139]:= a
b
c
d
Out[139]= 586
Out[140]= 679
Out[141]= 974
Out[142]= 4

In[115]:= a (2^10)^3 + b (2^10)^2 + c (2^10) + d

Out[115]= 629925689348

Idea is the same as the solution given:

# Python 2.7
from math import sqrt

lst= []
dct = {}

for a in range(1, 1025):
    if abs(sqrt(a)-int(sqrt(a))-0.208973068654) <= 0.00000001:
        lst.append(a)
        dct['c'] = a
    if abs(sqrt(a)-int(sqrt(a))-0.057628441591) <= 0.00000001:
        lst.append(a)
        dct['b'] = a
    if abs(sqrt(a)-int(sqrt(a))-0.207436873820) <= 0.00000001:
        lst.append(a)
        dct['a'] = a

for a in range(1,33):
    if a + sum(lst) == 2241:
        lst.append(a**2)
        dct['d'] = a**2
        break

print 1024**3*dct['a']+1024**2*dct['b']+1024*dct['c']+dct['d']

Subtracting equation 4 from equation 1 and equation 2 from equation 3, we get 2 equations : $a + \sqrt{d} - \sqrt{a} - d = 559.79256312618\\ c + \sqrt{b} - \sqrt{c} - b = 289.84865537294$

now check for solution within the range of the variables :

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53

//c++ 11
#include<iostream>
#include<cmath>

using namespace std;

int calc1(long , long );
int calc2(long , long );

int main()
{
    long a,b,c,d;
    long long ans=0;

    for(d=1;d<1025;d++)
        for(a=1;a<1025;a++)
            if(calc1(a,d))
                ans+=((pow(2,30)*a)+d);

    for(b=1;b<1025;b++)
        for(c=1;c<1025;c++)
            if(calc2(b,c))
                ans+=((pow(2,20)*b)+(pow(2,10)*c));


    cout.setf(ios::fixed);
    cout<<ans;

    return 0;
}

int calc1(long a, long d)
{
    double n1;

    n1=a-sqrt(a)-d+sqrt(d);

    if (floor(1000*n1)==559792)
        return 1;

    return 0;
}

int calc2(long b, long c)
{
    double n2;
    n2=c-sqrt(c)-b+sqrt(b);

    if (floor(1000*n2)==289848)
        return 1;

    return 0;
}

Output :

1
2
3
4
5

629925689348

Process returned 0 (0x0)   execution time : 0.742 s

Press any key to continue.

P.S. Ivan's solution is much better, mine just shows how lazy i am ;)