Make the year with cubes!

The sum of cubes $a^3+b^3$ can be factored as $(a+b)(a^2-ab+b^2)$ . If there exist integer solutions for $a$ and $b$ , then $(a+b)$ and $(a^2-ab+b^2)$ must be factors of $2015$ .

Let $m=a+b$ and $n=a^2-ab+b^2$ . Solving this system of equations for $a$ and $b$ yields $a=\frac{3m\pm\sqrt{12n-3m^2}}{6}$ and $b=m-a$ .

The possible factor pairs for $2015$ are: $1\times2015$ , $5\times403$ , $13\times155$ , and $31\times65$ .

Substituting these factor pairs of $2015$ into $m$ and $n$ yields only two possible integer solutions for $a$ and $b$ : $a=14$ and $b=-9$ , or $a=-9$ and $b=14$ . In either case, $a+b=\boxed{5}$ .

We make four observations:

1. $a+b>0$ ,

2. $a^3+b^3\equiv2\pmod{3}$ , so that $a+b\equiv2\pmod{3}$ ,

3. $2015=a^3+b^3\geq{\frac{1}{4}(a+b)^3}$ for $a+b>0$ , so that $a+b\leq{20}$ , and

4. $a^3+b^3=(a+b)(a^2-ab+b^2)$ , so that $a+b$ divides $2015=5\times13\times31.$

This leaves the possibility $a+b=\boxed{5}$ only.

I know this doesn't prove anything at all (unlike other better solutions), but I wanted to share how I solved this problem computationally with this little program in Ruby (even if you don't know Ruby it isn't hard to understand):

a = b = 0
for i in -28..28 #edit
    for j in -28..28
        a = i
        b = j
        result = a**3 + b**3
        puts a + b if result == 2015
    end
end

#outputs:
#=> 5
#=> 5

Again, I guess this doesn't have mathematical rigor, but I found the answer quickly.

@edit: Actually, we can show that a and b must be both in the range [-27, 27] since 27³ - 26³ = 2107 > 2015. If we try a's and b's bigger than 27, no matter the combination we can't get 2015 with a³ + b³ (because it will be bigger). This solution uses simple brute force, but given this constraint it executes pretty fast (therefore, we can see that this implementation efficiency deppends on the size of the equation's right side).

We can even extend it to any right side number with:

a = b = 0
right_number = 2015
limit = 1
limit += 1 while (limit**3 - (limit - 1)**3) < right_number
for i in -limit..limit
    for j in -limit..limit
        a = i
        b = j
        result = a**3 + b**3
        puts a + b if result == right_number
    end
end

$a^3+b^3=(a+b)(a^2-ab+b^2)=2015$

So $a+b$ must be a factor of 2015: 1, 5, 13, 31, 5·13, 5·31, 13·31, 5·13·31= 2015. We can rule out 2015 as $a^3+b^3>a+b$ here (see comment below).

Modulus 7: $\forall x, x^3\equiv 0$ , $x^3\equiv 1$ or $x^3\equiv -1$ . $a^3+b^3=2015\equiv -1 \Rightarrow a^3\equiv 0, (b^3\equiv -1\leftrightarrow b\equiv3,5,6)$ or the opposite. Then, $a+b\equiv 3,5,6$ and that leaves only $a+b=$ 5, 13 or 31.

Modulus 3: $\forall x, x^3\equiv x$ . So $a^3+b^3 \equiv a+b \equiv 2015 \equiv 2$ . That leaves only $\boxed{a+b=5}$ .

If we want $a$ and $b$ (for example to prove a solution really exists!), we solve $a^3+(5-a)^3=2015 \Leftrightarrow 15a^2-75a+125=2015$ $\Leftrightarrow 15(a-14)(a+9)=0$ . And we find 14 and -9.

$a^3+b^3>a+b$ :

As 2015 is not a cube, $a$ and $b$ can't be 0. As 2014 and 2016 are not cubes, they can't be 1 either.

As $a^3+b^3=2015>0 \Rightarrow a^3>-b^3 \Leftrightarrow a>-b \Leftrightarrow a+b>0$ , then at most one of them can be negative.

Suppose, WLOG, that $a\geq b$ , so $a$ and $a-b$ are positive. $a^2-ab+b^2=a(a-b)+b^2\geq b^2 > 1$ .

Therefore $a^3+b^3=(a+b)(a^2-ab+b^2)>a+b$ .

2015 is the product of 5, 13 and 31. Since a and b are said to be integers, 5 could be one of the three answers, the other two being 13 and 31.

I actually didn't realise it could be negative numbers so I thought 13 would be maximum. Then I realise it couldn't work - with a little help from my trusty calculator. Eventually, through trial and error I found the answer.

${ a }^{ 3 }+{ b }^{ 3 }=\quad (a+b)({ a }^{ 2 }-ab+{ b }^{ 2 })=(a+b)(\quad { (a+b) }^{ 2 }-3ab\quad )=5*13*31[factors\quad of\quad 2015]\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad .\\ Let\quad (a+b)=F:\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad .\\ F({ F }^{ 2 }-3ab)=5*13*31\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad .\\ \quad ({ F }^{ 2 }-3ab)=\frac { 5*13*31 }{ F } \qquad \qquad \qquad \qquad \qquad \quad \quad 5\equiv 2(mod\quad 3)\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad \quad \quad \quad \quad 13\equiv 1(mod\quad 3)\\ { F }^{ 2 }-\frac { 5*13*31 }{ F } =-3ab\qquad \qquad \quad \quad \quad \quad \quad \qquad \nearrow \quad \quad 31\equiv 1(mod\quad 3)\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad .\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad .\\ \\ Since\quad right\quad side\quad is\quad integer,\quad F\quad is\quad a\quad factor\quad of\quad 2015.\quad Also,\quad \\ the\quad right\quad side\quad must\quad be\quad a\quad factor\quad of\quad 3:\\ \\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad .\\ \\ If\quad F\equiv 1(mod\quad 3)\quad \quad \quad \quad \quad \quad [meaning\quad F=13\quad or\quad 31]\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad .\\ { (1) }^{ 2 }-\frac { 2*1*1 }{ 1 } \equiv 2(mod\quad 3)\quad \quad \quad \quad \quad \therefore F\neq 13\quad or\quad 31\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad .\\ \\ If\quad F\equiv 2(mod\quad 3)\quad \quad \quad \quad \quad \quad [meaning\quad F=5]\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad .\\ { (2) }^{ 2 }-\frac { 2*1*1 }{ 2 } \equiv 1(mod\quad 3)\quad \quad \quad \quad \quad \therefore \quad F=5\\ \\$

We can easily notice that $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ . The expression can be simplified as following:

$(a+b)^3 = a^3 + b^3 + 3ab\cdot (a+b)$

$a^3 + b^3 = (a+b)\cdot [ (a+b)^2 -3ab]$

We shall denominate $S = a+b$ (as S stands for sum ) and $P=ab$ (as P stands for product ). Therefore:

$2015 = S\cdot (S^2-3P)$

Also, let $x=S$ and $y=S^2-3P$ , so $2015 = x\cdot y$ . By factoring the number 2015, we know that $2015 = 5 \cdot 13 \cdot 31$ . This leaves us with a total of 16 possibilities for the ordenate pair $(x,y)$ : $(1,2015),(2015,1),(-1,-2015),(-2015,1),(5,403),(403,5),(-5,-403),(-403,-5),(13,155),(155,13),(-13,-155),(-155,-13),(65,31),(31,65),(-65,-31)$ and finally $(-31,-65)$ . That's a lot of possibilities! We should make some further considerations in order to eliminate some of them, starting with all negative ones. As $a$ and $b$ are both integers, one could think $x= -5$ and $y= -403$ (for example) would be a valid solution; but we can perform a fast analysis that guarantees that $S > 0$ (or $x>0$ ), as follows:

Assuming $a+b \leq 0$ , then $a \leq -b$ . So $a^3 \leq -b^3$ and $a^3 + b^3 \leq 0$ , which is absurd. That means $a+b > 0$ and, consequently, both $x$ and $y$ must be positive integers.

If $a$ and $b$ are the solutions of an equation, then it is the following:

$z^2 -Sz + P = 0$

As both solutions are integers (and therefore real numbers), $S$ and $P$ must satisfy $S^2 - 4P \geq 0$ . In terms of $x$ and $y$ , this means $y \geq \frac {1}{4}x^2$ . In the end, considering the possibilities which satisfy this condition and simultaneously guarantee that both $S$ and $P$ are integers, the only ordenate pair that "passes the test" is $(x,y) = (5,403)$ (meaning $S=5, P=-126$ and $a,b) = (14,-9)$ or vice-versa).

We've developed the solution in order that our answer is $x$ (or $S$ ), so, therefore:

$a+b = 5$

Hope it helped!

(a+b)^3=a^3+b^3+3ab^2+3ba^2. =2015+3ab(a+b) so. (a+b).((a+b)^2-3ab)=2015=5×403 a+b=5 or 403 .assume a+b=403. So ((a+b)^2-3ab=5 .so. (403)^2-3ab=5 so ab=(162409-5)\3=54134.67). which it is not an integer number &from given. a&b integer so ab must be integer number so a+b=5#########

a^3 + b^3 = 2015 means that: (a + b)(a^2 - ab + b^2) = 2015, where both (a + b) and (a^2 - ab +b^2) are integers.

The only pairs of integer factors of 2015 are: 1 and 2015 5 and 403 13 and 155 31 and 65,

and their negatives. For each pair, let (a + b) equal one factor, and let (a^2 - ab + b^2) equal the other factor. This gives a nonlinear system of equations that can be solved for a and b. Solving the system for each choice of (a + b) shows that the only choice that gives integer solutions for a and b is...a + b =5.

The sum of cubes $a^3+b^3$ can be written as $(a+b)(a^2-ab+b^2)\\$ Due to the symmetry of the expression, we can consider the factor pairs of 2015: $\\\{(1,2015),(5,403),(13,155)(31,65)\}$ and, as $(a+b) < (a^2-2ab+b^2) \rightarrow (a+b) \in \{1,5,13,31\}\\$ Try $(a+b)=5\\$ Then $(a^2+2ab+b^2)=25\\$ Substracting $(a^2-ab+b^2)=403\\$ Gives $3ab=-378 \rightarrow ab=-126 \\$ The factor pairs of 126 are $\{(1,126),(2,63),(3,42),(6,21),(9,14)\}\\$ Given the assumption that $(a+b)=5$ we see that $a=14$ and $b=-9$ satisfies the balance of the equation hence $(a+b)=\boxed{5}$

a^3 + b^3 = 2015 means that: (a + b)(a^2 - ab + b^2) = 2015, where both (a + b) and (a^2 - ab +b^2) are integers.

The only pairs of integer factors of 2015 are: 1 and 2015 5 and 403 13 and 155 31 and 65

For each pair, let (a + b) equal one factor, and let (a^2 - ab + b^2) equal the other factor. This gives a nonlinear system of equations that can be solved for a and b. Solving the system for each choice of (a + b) shows that the only choice that gives integer solutions for a and b is...a + b =5.

a^3 + b^3 = 2015 means that: (a + b)(a^2 - ab + b^2) = 2015, where both (a + b) and (a^2 - ab +b^2) are integers.

The only pairs of integer factors of 2015 are: 1 and 2015 5 and 403 13 and 155 31 and 65

For each pair, let (a + b) equal one factor, and let (a^2 - ab + b^2) equal the other factor. This gives a nonlinear system of equations that can be solved for a and b. Solving the system for each choice of (a + b) shows that the only choice that gives integer solutions for a and b is...a + b =5.