Question

If ‘a’ and ‘b’ are positive integers such that ${{a}^{2}}-{{b}^{4}}=2009$, find a + b.

Answer 1

Hint: Use the algebraic identity ${{x}^{2}}-{{y}^{2}}=\left( x+y \right)\left( x-y \right)$ to simplify the L.H.S of the given expression. Now, use the prime factorization method to find all the factors of 2009 and write it as the product of its two factors to form various cases. Consider $\left( a+{{b}^{2}} \right)=m$ and $\left( a-{{b}^{2}} \right)=n$, find the solve for the values of b. Check the case in which ‘b’ will be an integer and use which to find the corresponding value of ‘a’ to get the answer.

Complete step-by-step solution:
Here we have been provided with the equation ${{a}^{2}}-{{b}^{4}}=2009$ with the condition that ‘a’ and ‘b’ are positive integers. We are asked to find the value of a + b. First we need to determine the value of ‘a’ and ‘b’.
Now, using the algebraic identity ${{x}^{2}}-{{y}^{2}}=\left( x+y \right)\left( x-y \right)$ for simplifying the expression in the L.H.S we get,
$\Rightarrow \left( a+{{b}^{2}} \right)\left( a-{{b}^{2}} \right)=2009$
Assuming $\left( a+{{b}^{2}} \right)=m$ and $\left( a-{{b}^{2}} \right)=n$ we have,
$\Rightarrow m\times n=2009$
Writing 2009 as the product of its prime factors we have $2009=7\times 7\times 41$, so we can form various cases where we will write 2009 as the product of its two factors and compare with the values of m and n. Note that m > n, so we have,
(1) $m\times n=2009\times 1$, so on comparing we have m = 2009 and n = 1, operating m – n we get,
$\begin{align}
  & \Rightarrow 2{{b}^{2}}=2008 \\
 & \Rightarrow {{b}^{2}}=1004 \\
\end{align}$
Taking positive square root both the sides we get,
$\Rightarrow b=\sqrt{1004}$
Clearly we can see that 1004 is not a perfect square so ‘b’ will not be an integer. Therefore the above case is not possible.
(2) $m\times n=287\times 7$, so on comparing we have m = 287 and n = 7, operating m – n we get,
$\begin{align}
  & \Rightarrow 2{{b}^{2}}=280 \\
 & \Rightarrow {{b}^{2}}=140 \\
\end{align}$
Taking positive square root both the sides we get,
$\Rightarrow b=\sqrt{140}$
Clearly we can see that 140 is not a perfect square so ‘b’ will not be an integer. Therefore the above case is not possible.
(3) $m\times n=49\times 41$, so on comparing we have m = 49 and n = 41, operating m – n we get,
$\begin{align}
  & \Rightarrow 2{{b}^{2}}=8 \\
 & \Rightarrow {{b}^{2}}=4 \\
\end{align}$
Taking positive square root both the sides we get,
$\begin{align}
  & \Rightarrow b=\sqrt{4} \\
 & \Rightarrow b=2 \\
\end{align}$
Clearly we have the value of ‘b’ a positive integer so substituting it in the expression m = 49 we get,
$\begin{align}
  & \Rightarrow a=49-4 \\
 & \Rightarrow a=45 \\
\end{align}$
Hence, the value of a + b = 45 + 2 = 47.

Note: Note that in the first case we haven’t considered m = 1, n = 2009 and similarly in the second and third case we haven’t considered m = 7, n = 287 and m = 49, n = 41 respectively because $a+{{b}^{2}}$ will always be greater than $a-{{b}^{2}}$ and if we will consider the above three cases then the value of b turn out to be imaginary which is not possible.