Waste less time on Facebook — follow Brilliant.
×

Set of interesting problems.

These are some really cool problems. Try them!

  • \(\displaystyle \dfrac{1}{x} + \dfrac{1}{y}=\dfrac{1}{n}\) 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

Note by Jordi Bosch
3 years, 2 months ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

For 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

Log in to reply

@Patrick Corn Wau! I solved it in a much really complicated way!. Thanks! Jordi Bosch · 3 years, 2 months ago

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

Log in to reply

@Bogdan Simeonov Your conclusion also proves that \((x,y)=(0,0)\), the trivial solution, is the only solution over all integers. Daniel Liu · 3 years, 2 months ago

Log in to reply

@Bogdan Simeonov Nice idea! This is the way I solved it:

  • Consider \(gcd(x,y)=1\) because if they shared any factor we could divide the whole equation by it and it would remain identical. Now we see the RHS is multiple of 3, so the LHS must also. Since quartics are quadratics also and we know they are congruent with\( 0, 1\) modulo\( 3\). We get the LHS must be congruent with 0 modulo 3. But that's impossible because it' d mean x and y are congruent 0 modulo 3 and so they both are multiples of 3 contradicting the fact that \(gcd(x,y)=1\)
Jordi Bosch · 3 years, 2 months ago

Log in to reply

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

Log in to reply

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

Log in to reply

@Daniel Liu Same solution as mine! you nailed it! Jordi Bosch · 3 years, 2 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...