Question

Determine all pairs of positive integers $\left( {m,n} \right)$ for which ${2^m} + {3^n}$ is a perfect square.

Answer 1

Hint: Given the expression and we have to find the ordered pairs which satisfy the given condition. First, we will determine the value of one variable in terms of another variable. Then we will determine the value of the expression using the algebraic identity. Then find the value of the variables using substitution.

Complete step by step answer:
In the expression let us assume that the expression, ${2^m} + {3^n} = {k^2}$ is a perfect square.

Now, we will determine the value of ${3^n}$ by subtracting ${2^m}$ from both sides.

\[ \Rightarrow {2^m} + {3^n} - {2^m} = {k^2} - {2^m}\]

\[ \Rightarrow {3^n} = {k^2} - {2^m}\]

Now, assume \[m\] is even which is equal to \[2p\]

Now, we will substitute \[m = 2p\] into the value of \[{3^n}\]

\[ \Rightarrow {3^n} = {k^2} - {2^{2p}}\]

On rewriting the expression, we get:

\[ \Rightarrow {3^n} = {k^2} - {\left( {{2^p}} \right)^2}\]

Now, we will apply the algebraic identity \[{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)\] to the RHS of the expression.

\[ \Rightarrow 1 \times {3^n} = \left( {k - {2^p}} \right)\left( {k + {2^p}} \right)\]

Now, on comparing the expression on LHS to the expression on RHS we get:

\[k - {2^p} = 1\] …. (1)
\[k + {2^p} = {3^n}\] …. (2)

Now, subtract the equation (1) from equation (2).

\[ \Rightarrow k + {2^p} - \left( {k - {2^p}} \right) = {3^n} - 1\]

On simplifying the expression, we get:

\[ \Rightarrow k + {2^p} - k + {2^p} = {3^n} - 1\]

\[ \Rightarrow {2^p} + {2^p} + 1 = {3^n}\]

On rewriting the expression, we get:

\[ \Rightarrow 2 \times {2^p} + 1 = {3^n}\]

Now we will apply the algebraic identity \[{x^a} \cdot {x^b} = {x^{a + b}}\] to the expression.

\[ \Rightarrow {2^{p + 1}} + 1 = {3^n}\]

When the number is represented as power of two, the number will give the remainder \[4\], which means the value of \[n\] is even.

\[ \Rightarrow {\left( { - 1} \right)^n} \equiv {3^n}\left( {\bmod 4} \right)\] and \[ \Rightarrow {2^{p + 1}} \equiv 1\]

Let us assume \[n = 2q\]. Now, we will substitute \[n = 2q\] into the expression \[{2^{p + 1}} + 1 = {3^n}\].

\[ \Rightarrow {2^{p + 1}} + 1 = {3^{2q}}\]

On rewriting the expression, we get:

\[ \Rightarrow {2^{p + 1}} = {\left( {{3^q}} \right)^2} - {1^2}\]

Now, we will apply the algebraic identity \[{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)\] to the RHS of the expression.

\[ \Rightarrow 2 \times {2^p} = \left( {{3^q} - 1} \right)\left( {{3^q} + 1} \right)\]

Now, on comparing the expression on LHS to the expression on RHS we get:

\[ \Rightarrow {3^q} - 1 = 2\] …. (3)

On simplifying the expression, we get:

\[ \Rightarrow {3^q} = 2 + 1\]

\[ \Rightarrow {3^q} = {3^1}\]

Now, we will have the property of exponents, if \[{a^b} = {a^c}\] then \[b = c\].

Therefore, \[q = 1\]

Now, we will find the value of \[p\] by substituting the value of \[q\] into the equation \[{2^{p + 1}} + 1 = {3^{2q}}\].

\[ \Rightarrow {2^{p + 1}} + 1 = {3^{2 \times 1}}\]

On further solving the expression, we get:

\[ \Rightarrow 2 \times {2^p} = {3^2} - 1\]

\[ \Rightarrow 2 \times {2^p} = 8\]

Now, divide both sides by \[2\].

\[ \Rightarrow {2^p} = 4\]

Now, write the value at RHS as a power of \[2\]

\[ \Rightarrow {2^p} = {2^2}\]

Now, we will have the property of exponents, if \[{a^b} = {a^c}\] then \[b = c\].

Therefore, \[p = 2\]

Now, compute the value of \[m\] by substituting the value into the equation \[m = 2p\].

\[ \Rightarrow m = 2 \times 2\]

\[ \Rightarrow m = 4\]

Now, compute the value of \[n\] by substituting the value into the equation \[n = 2q\].

\[ \Rightarrow n = 2 \times 1\]

\[ \Rightarrow n = 2\]

Therefore, the value of \[m = 4\] and \[n = 2\]

Hence the pair of positive integers $\left( {4,2} \right)$.

Note: In such types of questions students mainly get confused in applying the formula. As they don't know which formula they have to apply. So, in such types of questions students may get confused in applying the algebraic identity to the expression. Also, students can make mistakes while applying the property of exponents in such types of questions.