Consecutive positive integers proof problem

Question

Consider any three consecutive positive integers. Prove that the cube of the largest cannot be the sum of the cubes of the other two.

Work: I tried to prove via contradiction.

I made three integers, k, k+1, and k+2. Then set the equation$(k+2)^3 = (k+1)^3 + k^3$

This equation is then expanded to $k^3+6k^2+12k+8 = 2k^3+3k^2+3k+1$ Ultimately, I get to a dead end $k^3-3k^2+9k = 7$.

I do not know where to go from this and it just seems that I took the wrong route in the beginning. I am currently learning about linear combinations and the division algorithm as well as Euclidean Algorithm, but I do not see anyway I can use those on this problem.

Answer 1

You have a cubic (actually, you made an arithmetic error, it should be $k^3 - 3k^2 -9k = 7$); rewrite it as $k^3-3k^2-9k-7=0$. You want integer roots for the cubic. Any integer roots must be factors of $7$, so there are only four possibilities, and checking them, there are no roots.

By the way, the arithmetic is slightly easier if you use $k+1$, $k$, and $k-1$.

Answer 2

Small trick: Let us consider $3$ integers of the form, $(a-1), a, (a + 1)$ and now suppose the cube of the largest is the sum of the cubes of the other two.

$(a + 1)^3 = a^3 + (a - 1)^3 \implies a^3 + 3a^2 + 3a + 1 = 2a^3 - 3a^2 + 3a - 1\implies a^3 - 6a^2 - 2 = 0 \implies a^2(a - 6) = 2$

From here, you can get to a contradiction in a million ways. Here's one:-

Now $a^2 \ge 0$ so for the product to be positive $(a - 6) \ge 0 \implies a \ge 6$. But then $a^2\ge 36$. The two products can never be equal to $2$ leading to a contradiction.