Find three consecutive positive integers a,b and c such that a+b is a perfect square and b+c is a perfect cube.

Question

Find three consecutive positive integers a,b and c such that a+b is a perfect square and b+c is a perfect cube.

Any ideas on how to begin?

Answer 1

$b=a+1$ and $c=a+2$. Write: $$2a+1=m^2,\ 2a+3=n^3$$ $$m^2+2=n^3$$ We need to look for a square that is 2 less than a cube. The squares are: $$1, 4, 9, 16, \mathbf{25}, 36, \dots$$ The cubes are: $$1, 8, \mathbf{27}, 64, 125, \dots$$ We see that $m^2=25$ satisfies the condition, so that $a=(25-1)/2=12$. Hence the consecutive integers with the given properties are 12, 13 and 14.

Answer 2

Probably one of the easiest things for you to do would be to search odd cube as b+c and check whether they fit the pattern. Example: the second odd cube is 27. If b = 13 and c = 14, then a = 12 and a + b = 25, which is a perfect square, so the triplet (12, 13, 14) fits the pattern. Based on the final digit of cubes and squares, the last digit of any cubes worth checking should end in 1, 3 or 7, as these are the only odd last digits that could fit the square condition.

Answer 3

For all $x,y \in \mathbb{N}$, $x^3-y^2=2$ according to the given question, where $x=a+b$ and $y=b+c$. However, $x^3 \equiv 0,\pm1 \pmod4$ and $y^2 \equiv 0,1 \pmod4$.

So to satisfy the given equation, the following must hold: $$x^3-y^2\equiv 2 \pmod4$$

For this means that the required $x,y$ are such that $x^3 \equiv -1\equiv 3 \pmod4$ and $y^2 \equiv 1 \pmod4$. That is, the required forms are $$x^3=4n+3$$ $$y^2=4n+1$$

See if this helps.