Find all the pairs

Find all the pairs

Postby Guest » Fri Dec 01, 2023 4:11 pm

Find all ordered pairs $(m,n) ∈ N^2$ for which $4^m + 5^n$ is a perfect square.
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Re: Find all the pairs

Postby Guest » Sat Mar 30, 2024 7:49 am

To find all ordered pairs (m, n) where m and n are natural numbers (denoted as N) such that 4^m+5^n is a perfect square, we need to explore the possible values systematically.

Let's break down the problem:

Understanding Perfect Squares: A perfect square is a number that can be expressed as the product of an integer with itself. For instance, 4, 9, 16, etc., are perfect squares.
Given Expression: We're given the expression 4^m+5^n.

Finding Ordered Pairs (m, n): We're tasked with finding all pairs (m, n) such that 4^m+5^n results in a perfect square.

To approach this problem:

We need to think about the properties of perfect squares and the nature of the given expression.

Since both 4m and 5n are raised to powers, the resulting sum can vary widely.

We must consider that perfect squares are usually formed by adding two numbers that are either both squares themselves or such that their sum results in a square.

Let's analyze the cases:

Both 4m and 5n are perfect squares:

If both terms are perfect squares, then 4^m+5^n is a sum of two perfect squares, which may or may not result in another perfect square depending on their values.
One term is a perfect square, and the other isn't:

If one term is a perfect square, say 4^m, and the other isn't (5^n), then we have to consider whether their sum can result in a perfect square.
Both terms are not perfect squares:

In this case, it might be more challenging to find solutions, as we would have to examine various combinations of powers of 4 and 5 to see if their sum results in a perfect square.

Overall, solving this problem involves a systematic exploration of the powers of 4 and 5, checking if their sum leads to a perfect square, and then identifying all possible ordered pairs (m, n) that satisfy the condition.

By the way, if you need further assistance with math assignments or want to explore more problems, you might find useful resources at MathsAssignmentHelp.com. You can contact them at +1 (315) 557-6473.
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Re: Find all the pairs

Postby Guest » Sat Apr 06, 2024 7:43 am

To find all ordered pairs (m, n) ∈ ℕ^2 for which 4^m+5^n is a perfect square, let's denote the perfect square as k^2, where k is a positive integer.
So, we have: 4^m+5^n =k^2
Now, let's consider the cases where m and n can be:
Case 1: If m=0:
4^0+5^n=1+5^n
For 1+5^n to be a perfect square, it implies that n must be 0. So, for this case, we have the solution (0, 0).
Case 2: If n=0:
4^m+5^0=4^m+1
For 4^m+1 to be a perfect square, 4^m must be one less than a perfect square. This happens only when m=0. So, for this case, we also have the solution (0, 0).
Case 3: If m>0 and n>0: For this case, we rewrite the equation as:
4^m=(k−5^n)(k+5^n)
Since 4^m is a power of 2, k−5^n and k+5^n must both be powers of 2 as well.
Let k−5^n=2x and k+5^n=2^y, where x<y.
Adding these two equations, we get:
2k=2^x+2^y
k=2^x−1+2^y−1
Now, subtracting the equations, we get: 2⋅5^n=2^y−2^x
But 2⋅5^n can never be a power of 2, hence contradiction.
Thus, the only solutions are (0, 0).

By the way, if you need further assistance with math assignments or want to explore more problems, you might find useful resources at website of Maths Assignment Help. You can contact them at +1 (315) 557-6473.
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