Question

Let $D = {a^2} + {b^2} + {c^2}$, where a and b are consecutive integers and c=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:The above question is based on the concept of solving mathematical equations. The main approach towards solving the above question is to bring the above equation in terms of the variable a by using c=ab and also using consecutive integers to form an equation so that we can find the value of \[\sqrt D \]

Complete step by step solution:
In the above question there are three variables a, b and c which is written in the equation $D = {a^2} + {b^2} + {c^2}$ where a and b are consecutive integers.
For example, if the variable a is 1 then the next variable b is 2 since these two numbers are consecutive.
It can be written as,
\[b = a + 1\]
Given is c=ab. We can write it in terms of a by substituting the above equation in it.
\[c = ab = a\left( {a + 1} \right)\]
Therefore, substituting the above values in the main equation so that the whole equation is in terms of a.
$
\Rightarrow D = {a^2} + {b^2} + {c^2} \\
\Rightarrow D = {a^2} + {\left( {a + 1} \right)^2} + \left( {a{{\left( {a + 1} \right)}^2}} \right) \\
\Rightarrow D = {a^4} + 2{a^3} + 3{a^2} + 2a + 1 \\
\Rightarrow D = {\left( {{a^2} + a + 1} \right)^2} \\
\Rightarrow \sqrt D = \left( {{a^2} + a + 1} \right) \\
$
Now if a is odd then we get the outcome as odd and is the value of a is even then we get the value as odd.
Therefore option B is correct.

Note: An important thing to note is that when we get the above result, we can check by substituting whether it is odd or even. For example, a=1 which is odd therefore \[\left( {{1^2} + 1 + 1} \right) = 3\] and a=2 which is even therefore \[\left( {{2^2} + 2 + 1} \right) = 7\]which still gives odd.