Set of interesting problems.

These are some really cool problems. Try them!

  • 1x+1y=1n\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 x4+y4=3x3yx^{4} + y^{4} = 3x^{3}y

  • Find the solutions of the following system of equations:

xyz=42x - \sqrt{yz} = 42

yxz=6y - \sqrt{xz} = 6

zxy=30z - \sqrt{xy} = 30

Here's the second part

Note by Jordi Bosch
6 years ago

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

Patrick Corn - 6 years ago

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

Jordi Bosch - 6 years ago

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For the second one:

Since 33x3y 3|3x^3y ,then 3x4+y43|x^4+y^4.That means that 3|x and 3|y.

Let x and y be the smallest integer solutions.From before, x=3x1x=3x_1 and y=3y1y=3y_1

So x14+y14=3x13y1x_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 - 6 years ago

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

Daniel Liu - 6 years ago

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Nice idea! This is the way I solved it:

  • Consider gcd(x,y)=1gcd(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 with0,1 0, 1 modulo3 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)=1gcd(x,y)=1

Jordi Bosch - 6 years ago

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Here is the second part. Follow me if you want more problems!

Jordi Bosch - 6 years ago

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First one:

Clearing denominators gives xy=nx+nyxy=nx+ny or xynxny=0xy-nx-ny=0.

Using Simon's Favorite Factoring Trick, we get (xn)(yn)=n2(x-n)(y-n)=n^2.

We are given that this has 505505 ordered pairs of positive integers that satisfy it. Note that x,ynx,y \ge n or else one of them is negative. Thus, xnx-n and yny-n are both positive, so we just need to equate the number of factors of n2n^2 to 505505.

We have a few cases now:

n2=p504n^2=p^{504} for a prime pp. Clearly, n=p252n=p^{252}, so nn is a perfect square.

n2=p4q100n^2=p^4\cdot q^{100} for primes p,qp,q. Clearly, n=p2q50n=p^2\cdot q^{50}, so nn is a perfect square.


Daniel Liu - 6 years ago

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Same solution as mine! you nailed it!

Jordi Bosch - 6 years ago

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