Question

Three consecutive positive integers are raised to the first, second and the third powers respectively and then added. The sum so obtained is a perfect square whose square root is equal to the total of the three original integers. Which of the following best describes the minimum say n of these three integers?
[a] $1\le n\le 3$
[b] $4\le n\le 6$
[c] $7\le n\le 9$
[d] $10\le n\le 12$

Answer 1

Hint:Use the fact that three consecutive integers can be written as n, n+1,n+2 where n is the smallest of the integers. Hence find the sum of first, second and third number raised to the first, second and third powers respectively. Equate this to the square of the sum of the three integers and hence form a cubic in n. Solve for n. Hence find the range which n satisfies.

Complete step-by-step answer:
Let the three integers be n,n+1 and n+2.
Now, we have sum of first integer raised to the first power, second integer raised to the second power and the third integer raised to the third power is given by
$S=\left( n \right)+{{\left( n+1 \right)}^{2}}+{{\left( n+2 \right)}^{3}}$
But according to the question $S={{\left( n+n+1+n+2 \right)}^{2}}$
Hence, we have
$\begin{align}
  & n+{{\left( n+1 \right)}^{2}}+{{\left( n+2 \right)}^{3}}={{\left( n+n+1+n+2 \right)}^{2}} \\
 & \Rightarrow n+{{\left( n+1 \right)}^{2}}+{{\left( n+2 \right)}^{3}}={{\left( 3n+3 \right)}^{2}} \\
\end{align}$
Taking 3 common from the terms of 3n+3, we get
$n+{{\left( n+1 \right)}^{2}}+{{\left( n+2 \right)}^{3}}=9{{\left( n+1 \right)}^{2}}$
Subtracting ${{\left( n+1 \right)}^{2}}$ from both sides, we get
$n+{{\left( n+2 \right)}^{3}}=8{{\left( n+1 \right)}^{2}}$
Using ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}},$ we get
$n+{{\left( n+2 \right)}^{3}}=8\left( {{n}^{2}}+2n+1 \right)$
Using ${{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}},$ we get
$n+{{n}^{3}}+6{{n}^{2}}+12n+8=8{{n}^{2}}+16n+8$
Transposing all the terms to LHS, we get
$\begin{align}
  & {{n}^{3}}+6{{n}^{2}}-8{{n}^{2}}+12n+n-16n+8-8=0 \\
 & \Rightarrow {{n}^{3}}-2{{n}^{2}}-3n=0 \\
\end{align}$
Taking n common from the term on LHS, we get
$n\left( {{n}^{2}}-2n-3 \right)=0$
Since n is a positive integer, we have $n\ne 0$
Hence, we have
${{n}^{2}}-2n-3=0$
Using splitting the middle term, we have
${{n}^{2}}-3n+n-3=0$
Taking n common from the first two terms and 1 common from the last two terms, we get
$n\left( n-3 \right)+1\left( n-3 \right)=0$
Taking n-3 common from the terms, we get
$\left( n-3 \right)\left( n+1 \right)=0$
Since n is a positive integer, we have $n+1\ne 0$
Hence, we have
$n-3=0$
Adding 3 on both sides, we get
$n=3$
Hence the smallest integer among the three consecutive integers is 3
Hence, we have
$1\le n\le 3$
Hence option [a] is correct.

Note: Verification:
We have the three integers are 3,4.5.
Now, we have
$3+{{4}^{2}}+{{5}^{3}}=3+16+125=144$
Also, we have
${{\left( 3+4+5 \right)}^{2}}={{12}^{2}}=144$
Hence we have the sum of the first power of the first number, the second power of the second number and the third power of the third number is a perfect square whose square root is equal to the sum of the three integers.
Hence our answer is correct.