How would you go about solving a problem like this:
Let a, b, and c be positive integers which satisfy $a^3+b^3+c^2=2012$. Find $a+b+c.$
It doesn't appear that there's enough information to solve if, but test is multiple choice and the answer provided are:
A. 28 B. 30 C. 32 D. 34 E. 36
The following is a paper and pencil method to solve the problem without a calculator.
Let $n$ be one of the numbers from $\{28,30,32,34,36\}$. We know that $$a^3+b^3=(a+b)(a^2-ab+b^2)=2012-c^2 \tag{1}$$ and that $$a+b=n-c \tag{2}$$
From this follows that $a$ and $b$ are solutions of
$$3x^2-3(n-c)x+(n-c)^2-\frac{2012-c^2}{n-c}=0 \tag{3}$$
A necessary condition that an integer solution for $(3)$ exists is $$n-c \mid 2012-c^2 \tag{4}$$.
We can assume
$$n-c \gt 8 \tag{5}$$
otherwise
$$a^3+b^3+c^2 \le (a+b)^3+c^2 \le (n-c)^3+n^2 \le 8^3 + 36^2 <2012$$
$n-c$ is a divisor of $2012-c^2$ and therefore a divisor of $$f(n)=\frac{n}{(2012,n)}(2012-c^2)-(\frac{2012}{(2012,n)}+\frac{n}{(2012,n)}c)(n-c)$$
$(a,b)$ is the greatest common divisor of $a$ and $b$.
If define $s(m,k)$ as the largest divisor of $m$ that does not contain a prime factor larger than $k$. We have $$d \mid m \; \text{and} \; d \le k \implies d \mid s(m,k) \tag{6}$$
So we get $$n-c \mid s(f(n),n) \tag{7}$$ The table lists the values of $f(n)$ and $s(f(n),n)$
$$ \tag{8} \begin{array}{r|r|r} n&f(n)&s(f(n),n) \\ \hline{} \\ 28 & 307 \cdot c & c\\ 30 & 2^2 139 \cdot c & 2^2 c\\ 32 & 13 \cdot 19 \cdot c & 13 \cdot 19 \cdot c\\ 34 & 2^2 107 \cdot c & 2^2 c\\ 36 & 179 \cdot c & c \end{array} $$
For $n \in \{28,36\}$ we get from $(4)$, $(6)$ and $(8)$
$$n-c \mid s((2012-c^2)+c \cdot c,n)$$
But $s(2012,n)=4$ and $(5)$ shows, there is no solution for $n=28$ and $n=36$.
For $n \in \{30,34\}$ we get from $(4)$, $(6)$ and $(8)$ $$n-c \mid s(4 \cdot (2012-c^2)+c \cdot (4c),n)$$ $s(4 \cdot 2012,n)=16$ and $(5)$ shows, the only possible value is $n-c=16$. But $16 \nmid c$, neither for $n=30, c=14$ nor for $n=34, c=20$. So there is no solution for $n \in \{30,34\}$.
$n=32$:
We get $32-c \mid 13 \cdot 19 \cdot c$. Here we have three cases:
Case 1 gives no solution for the same reasons as for $n=28$ and $n=36$.
Case 2 gives $32-c=13$ or $32-c=26$. therefore $c=13$. Substituting this values in $(1)$ gives the equations $$ x^2-13x+14=0$$ and $$x^2-26x+200=0$$ None of them has integer solutions.
Finally for case 3 we get $32-c=19$ and the equation
$$x^2-19x+88=0$$
with the two solutions $8$ and $11$.
So the solution to the problem is
$$8^3+11^3+13^2=2012$$ $$8+11+13=32$$
I'd proceed by guessing. I know that $8^3=512$, so that's a good place to start. Then we'd have $b^3+c^2=1500$. Then I'm going to guess $b=11$, so we have $b^3=1331$. This yields $c^2=169$, which fortunately has a solution of $c=13$. This makes the answer 32.
There may be a better way to do this, but the very limited number of solutions to $a,b$ makes guessing reasonably easy.
Although calculators I assume are not allowed, this is still quite easily guessed from trial and error. My advice to you would first work with large (a, b) and see if the resulting c will satisfy the equation.
The correct answer is 32 (8, 11, 13)
Since $2012 = 2 * 2 * 503$ and $8 | (2n)^3$, $a, b, c$ have to be made up of $2$ evens and $1$ odd. Case checking with a calculator confirms that this does not hold for any $(a, b, c)$.
– MT_ Oct 08 '13 at 21:49