Question

Let \[D{\text{ }} = {\text{ }}{a^2} + {\text{ }}{b^2} + {\text{ }}{c^2},\] where a and b are consecutive integers and \[c{\text{ }} = {\text{ }}ab.\] Then $ \sqrt D $ is.
A) always an even integer
B) always an odd integer
C) sometimes an odd integer, sometimes not
D) sometimes a rational number, sometimes not

Answer 1

Hint: To solve this problem, i.e., to find the value of $ \sqrt D , $ we will first assume the value of a, now since a and b are consecutive integer, we will get the value of b in terms of n, and then since, c is the product of a and b, we will get the value of c also in terms of n. Now on putting the values of a, b and c, we will get the value of $ \sqrt D . $ Afterwards, we will put the value of n, and then we will check, with the four multiple choices options, which will satisfy our answer.

Complete step-by-step answer:
We have been given that \[D{\text{ }} = {\text{ }}{a^2} + {\text{ }}{b^2} + {\text{ }}{c^2},\] where a and b are consecutive integers and \[c{\text{ }} = {\text{ }}ab.\] We need to find the value of $ \sqrt D $ .
So, it is given that \[D{\text{ }} = {\text{ }}{a^2} + {\text{ }}{b^2} + {\text{ }}{c^2}......eq.(1)\]
Now, let \[a{\text{ }} = {\text{ }}n,\] and since, a and b are consecutive integers, therefore,\[b = a + 1{\text{ }} = {\text{ }}n + {\text{ }}1\]and thus, \[c = ab = \left( n \right)\left( {n + 1} \right){\text{ }} = {\text{ }}{n^2} + n\]
On putting the values of \[a{\text{ }} = {\text{ }}n,{\text{ }}b = {\text{ }}n + {\text{ }}1\;\] and \[c = \left( {{n^2} + n} \right)\] in \[eq.{\text{ }}\left( 1 \right),\] we get
\[
D = {n^2} + {\left( {n + 1} \right)^2} + {\left( {{n^2} + n} \right)^2} \\
D = {n^2} + {n^2} + 2n + {1^2} + {n^4} + 2{n^3} + {n^2} \\
D = {n^4} + 2{n^3} + 3{n^2} + 2n + 1 \\
D = {({n^2} + n + 1)^2} \\
\Rightarrow \sqrt D = {n^2} + n + 1 \\
 \]
Now, let us check the values of $ \sqrt {D.} $
If \[n{\text{ }} = {\text{ }}1,\] then $ \sqrt D = {1^2} + 1 + 1 = 3 $
If \[n{\text{ }} = - 1,\] then $ \sqrt D = {( - 1)^2} + ( - 1) + 1 = 1 $
If \[n{\text{ }} = {\text{ }}2,\] then \[\sqrt D = {(2)^2} + 2 + 1 = 4 + 2 + 1 = 7\]
So, for any value of n, $ \sqrt D $ is an odd integer.

So, the correct answer is “Option D”.

Note: Students should note that in these types of questions, always assume that value, which is easier to evaluate in the end. Always try to make your solution easier, not complex. And always try to make the equation in linear variable.