# Positive Integer Solutions of $(x^2+y^2)^n=(xy)^{2014}$

• MHB
• anemone
In summary, a positive integer solution is a set of values for the variables in an equation that result in a positive integer when substituted into the equation. There are an infinite number of positive integer solutions for the equation $(x^2+y^2)^n=(xy)^{2014}$, as any positive integer value for x and y can be raised to the 2014th power. The exponent n must be a multiple of 2 for this equation to have positive integer solutions, and x and y must be relatively prime. This equation can only have two variables, x and y, to have positive integer solutions.
anemone
Gold Member
MHB
POTW Director
Find all positive integers $n$ for which the equation $(x^2+y^2)^n=(xy)^{2014}$ has positive integer solutions.

anemone said:
Find all positive integers $n$ for which the equation $(x^2+y^2)^n=(xy)^{2014}$ has positive integer solutions.

I looked for solutions in which $x$ and $y$ are both equal to the same power of $2$.

So suppose that $x=y=2^k$. Then the equation becomes $\bigl(2^{2k+1}\bigr)^n = \bigl(2^{2k}\bigr)^{2014}$,
$2^{(2k+1)n} = 2^{4028k}$,
$(2k+1)n = 4028k = 4*19*53k$.
Since $k$ and $2k+1$ are co-prime, it follows that $2k+1$ must be an odd divisor of $4028$, namely $19$, $53$ or $19*53 = 1007$. The corresponding values of $k$ are $9$, $26$ and $503$. That gives these three solutions to the problem:

$x=y=2^9, \ n = 4*53*9 = 1908$,

$x=y=2^{26}, \ n = 4*19*26 = 1976$,

$x=y=2^{503}, \ n = 4*503 = 2012$.

That's as far as I can go. I believe that those three solutions should be the only ones, but I can't see how to prove that there are no others.

I will post the model solution below, I hope you and the readers will enjoy reading the solution.
Assume that $(x^2+y^2)^n=(xy)^{2014}$ holds for some positive integers $n,\,x$ and $y$. From $x^2+y^2\ge 2xy>xy$ it follows that $n<2014$. Let $d=\gcd(x,\,y)$ and $a=\dfrac{x}{d}$, $b=\dfrac{y}{d}$. Then $d^{2n}(a^2+b^2)^n=d^{2\cdot 2014}(ab)^{2014}$, which gives the equality $(a^2+b^2)^n=d^{2\cdot (2014-n)}(ab)^{2014}$. As $b$ divides the right side of this equality, $(a^2+b^2)^n$ is divisible by $b$ as well. But because $\gcd(a,\,b)=1$, also $\gcd(a^2,\,b)=1$ and $\gcd(a^2+b^2,\,b)=1$ whence $\gcd((a^2+b^2)^n,\,b)=1$, so the only possibility is $b=1$. Due to symmetry, also, $a=1$. The above equality now takes the form $2^n=d^{2\cdot(2014-n)}$. Therefore, $d=2^k$ for some $k$ and $n=2k\cdot(2014-n)$, whence $n\cdot(2k+1)=4k\cdot 1007$. Since $\gcd(2k+1,\,4)=1$ and $\gcd(2k+1,\,k)=1$, $2k+1$ divides 1007. As $1007=19\cdot 53$ and $n$ has to be positive, the possible values for $k$ are 9, 26 and 503, and the corresponding values of $n$ are 1908, 1976 and 2012.

## 1. What is the definition of a positive integer solution?

A positive integer solution is a set of values for the variables in an equation that result in a positive integer answer.

## 2. How many positive integer solutions are there for the equation $(x^2+y^2)^n=(xy)^{2014}$?

The number of positive integer solutions for this equation is infinite.

## 3. Can this equation have solutions where $x$ and $y$ are not integers?

Yes, it is possible for $x$ and $y$ to be non-integer values and still satisfy the equation, as long as the result is a positive integer.

## 4. What is the relationship between the powers $n$ and $2014$ in this equation?

The powers $n$ and $2014$ do not have a specific relationship in this equation. They are both simply used as exponents to represent the equation.

## 5. Are there any specific patterns or strategies for finding positive integer solutions to this equation?

Yes, there are some patterns and strategies that can be used to find positive integer solutions for this equation, such as factoring and using modular arithmetic. However, there is no guarantee that these strategies will work for all cases.

• General Math
Replies
1
Views
837
• General Math
Replies
1
Views
811
• General Math
Replies
1
Views
767
• General Math
Replies
15
Views
2K
• General Math
Replies
1
Views
904
• General Math
Replies
11
Views
1K
• General Math
Replies
1
Views
795
• General Math
Replies
1
Views
845
• General Math
Replies
4
Views
894
• General Math
Replies
22
Views
1K