Finding Max n in $4^{27}+4^{500}+4^\text{n}=\text {k}^2$

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In summary, the maximum value of n is 972, corresponding to a value of k that makes the expression a perfect square. This is achieved by setting n to be 945, which is also the greatest possible solution. Any larger value for n would result in a contradiction.
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Albert1
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$ 4^{27}+4^{500}+4^\text{n}=\text {k}^2 $

where n and k are positive integers ,please find max(n)
 
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Albert said:
$ 4^{27}+4^{500}+4^n=k^2 $

where n and k are positive integers ,please find max(n)
First, notice that $n$ must be quite large. The reason for that is that $4^{500} = \bigl(2^{500}\bigr)^2$ is a square. The next square after that is $\bigl(2^{500}+1\bigr)^2 = 4^{500} + 2^{501} + 1$. So we must have $4^{27}+4^n > 2^{501} > 4^{250}$, and it follows that $n$ must be at least $250$.

In particular, $n$ is certainly greater than 27. So let $m = n-27$. Then $ 4^{27}+4^{500}+4^n= 4^{27}\bigl(4^{473} + 4^m + 1\bigr)$. Since $4^{27} = \bigl(2^{27}\bigr)^2$ is a square, we want $4^{473} + 4^m + 1$ to be a square. You can find one solution to this by noticing that $\bigl(2\cdot 4^{236} + 1\bigr)^2 = 4^{473} + 4^{237} + 1$. Thus $m=237$ is a solution. The corresponding value for $n$ is $n=237+27 = 264$.

Pushing that idea a bit further, we have another solution: $\bigl(2\cdot 4^{472} + 1\bigr)^2 = 4^{945} + 4^{473} + 1$. That gives a bigger solution, $m=945$, corresponding to $\boxed{n= 972}$.

Now we want to show that $n=972$, or $m=945$, is the greatest possible solution. The reason for that is that if $x>945$ then $4^x + 4^{473}+1 > 4^x = \bigl(2^x\bigr)^2$. If $4^x + 4^{473}+1$ is a square, then it must be at least as big as $\bigl(2^x+1\bigr)^2$. But $\bigl(2^x+1\bigr)^2 = 4^x + 2^{x+1} + 1$. Therefore $4^{473} \geqslant 2^{x+1} > 2^{946} = 4^{473}$, which is a contradiction.
 
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Opalg :well done (Yes)
 

FAQ: Finding Max n in $4^{27}+4^{500}+4^\text{n}=\text {k}^2$

What is the value of n in the equation $4^{27}+4^{500}+4^\text{n}=\text {k}^2$?

The value of n cannot be determined without knowing the value of k. This equation is an example of a Diophantine equation, which involves finding integer solutions for variables. In order to solve for n, we would need to know the value of k and use mathematical techniques such as factoring or modular arithmetic to find the possible values of n.

Is there a maximum value for n in this equation?

Yes, there is a maximum value for n in this equation. Since n is an exponent in the term $4^n$, the value of n cannot be infinite. However, the specific maximum value will depend on the value of k and the precision of the calculation.

How can we determine the value of k in this equation?

In order to determine the value of k, we would need to know the values of n and the precision of the calculation. From there, we can use a calculator or mathematical software to solve for k. Alternatively, we can use algebraic techniques such as factoring or completing the square to solve for k.

What is the significance of this equation in the field of mathematics?

This equation is an example of a Diophantine equation, which has been studied for centuries by mathematicians. It also involves the concept of perfect squares and prime numbers, which are important in number theory. Solving this equation can also have practical applications, such as in cryptography and coding theory.

Are there any real-world applications of this equation?

While this equation may not have direct real-world applications, the concepts and techniques used to solve it have practical applications in fields such as computer science, cryptography, and coding theory. Additionally, understanding the properties of perfect squares and prime numbers can have practical applications in fields such as finance and statistics.

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