Problem
Two expressions that cannot both be cubes
For a natural number n, prove that n + 2 and n² + n + 1 cannot both be perfect cubes.
Let \(n\) be a natural number. Prove that the two numbers
\[n+2\qquad\text{and}\qquad n^2+n+1\]cannot both be perfect cubes.
The statement holds whether or not your convention includes \(0\) among the natural numbers.
Explore the solution
Suppose, for a contradiction, that
\[n+2=a^3,\qquad n^2+n+1=b^3\]for positive integers \(a,b\). Since \(n\geq 0\), we have \(a^3\geq 2\), so \(a\geq 2\).
Substituting \(n=a^3-2\) gives
\[\begin{aligned} b^3&=(a^3-2)^2+(a^3-2)+1\\ &=a^6-3a^3+3. \end{aligned}\]This number lies strictly between two consecutive cubes. First,
\[a^6-3a^3+3<a^6=(a^2)^3,\]because \(a\geq 2\). For the lower bound, subtract the preceding cube:
\[\begin{aligned} &(a^6-3a^3+3)-(a^2-1)^3\\ &\qquad=3a^4-3a^3-3a^2+4\\ &\qquad=3a^2(a^2-a-1)+4>0. \end{aligned}\]The last inequality follows from \(a\geq 2\), which gives \(a^2-a-1\geq 1\). Thus
\[(a^2-1)^3<b^3<(a^2)^3.\]Taking cube roots gives \(a^2-1<b<a^2\), impossible for an integer \(b\). Therefore the two expressions cannot both be cubes.