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\(\displaystyle \dfrac{1}{x} + \dfrac{1}{y}=\dfrac{1}{n}\) has 505 ordered pairs(x,y) of positive integers that satisfy the equation. Prove that n is a perferct square.

Find all integer solutions of \(x^{4} + y^{4} = 3x^{3}y\)

Find the solutions of the following system of equations:

\(x - \sqrt{yz} = 42\)

\(y - \sqrt{xz} = 6\)

\(z - \sqrt{xy} = 30\)

Here's the second part

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TopNewestFor the last one, there are no solutions. Suppose \( (x,y,z) \) was a solution, and note that \( x,y,z \) must all be positive (indeed, they are at least \( 42,6,\) and \( 30 \) respectively). The equations imply \( x > \sqrt{yz}, y > \sqrt{xz}, z > \sqrt{xy}. \) Multiplying all these gives \( xyz > xyz \), which is a contradiction. – Patrick Corn · 3 years, 2 months ago

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– Jordi Bosch · 3 years, 2 months ago

Wau! I solved it in a much really complicated way!. Thanks!Log in to reply

For the second one:

Since \( 3|3x^3y\) ,then \(3|x^4+y^4\).That means that 3|x and 3|y.

Let x and y be the smallest integer solutions.From before, \(x=3x_1\) and \(y=3y_1\)

So \(x_1^4+y_1^4=3x_1^3y_1\)

This is a solution clearly smaller than x and y which is a contradiction.Thus there are no solutions in positive integers. – Bogdan Simeonov · 3 years, 2 months ago

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– Daniel Liu · 3 years, 2 months ago

Your conclusion also proves that \((x,y)=(0,0)\), the trivial solution, is the only solution over all integers.Log in to reply

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Here is the second part. Follow me if you want more problems! – Jordi Bosch · 3 years, 2 months ago

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First one:

Clearing denominators gives \(xy=nx+ny\) or \(xy-nx-ny=0\).

Using Simon's Favorite Factoring Trick, we get \((x-n)(y-n)=n^2\).

We are given that this has \(505\) ordered pairs of positive integers that satisfy it. Note that \(x,y \ge n\) or else one of them is negative. Thus, \(x-n\) and \(y-n\) are both positive, so we just need to equate the number of factors of \(n^2\) to \(505\).

We have a few cases now:

\(n^2=p^{504}\) for a prime \(p\). Clearly, \(n=p^{252}\), so \(n\) is a perfect square.

\(n^2=p^4\cdot q^{100}\) for primes \(p,q\). Clearly, \(n=p^2\cdot q^{50}\), so \(n\) is a perfect square.

QED – Daniel Liu · 3 years, 2 months ago

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– Jordi Bosch · 3 years, 2 months ago

Same solution as mine! you nailed it!Log in to reply