Question

Prove that there is no positive integer n such that\({n^2} + {n^3} = 100\)

Short answer

There is no positive integer n such that\({n^2} + {n^3} = 100\).

Step-by-step solution

Introduction

Consider the equation,

\({n^2} + {n^3} = 100\)

Substitution for\(n = 1\)and\(n = 2\) in \({n^2} + {n^3} = 100\)

Substitute\(n = 1\)

\(\begin{aligned}{}{1^2} + {1^3} = 100\\2 \ne 100\end{aligned}\)

Substitute\(n = 2\)

\(\begin{aligned}{}{2^2} + {2^3} = 100\\4 + 8 = 100\\12 \ne 100\end{aligned}\)

Substitution for\(n = 3\)and\(n = 4\) in \({n^2} + {n^3} = 100\)

Substitute\(n = 3\)

\(\begin{aligned}{}{3^2} + {3^3} = 100\\9 + 27 = 100\\36 \ne 100\end{aligned}\)

Substitute\(n = 4\)

\(\begin{aligned}{}{4^2} + {4^3} = 100\\16 + 64 = 100\\80 \ne 100\end{aligned}\)

For\(n > 4\),\({n^3} > 100\).

Hence, \({n^2} + {n^3} > 100\forall n > 4\).

Therefore, there is no positive integer n such that \({n^2} + {n^3} = 100\).