I think 0.
If a = 2x and b = 3x then a = 2/3 * b
So it would mean that root(2/3 * b) needs to be an integer.
root(b) is an integer (cause it's a perfect square) but there is no integer that multiplied by an irrational number is also an integer.
I'm not sure tho.

@Sam Segers
–
He answered it in a different context.
Since last week the same question was asked on brilliant, and he assumed it asked integers and not positive. Just let it go.

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last week algebra level 3

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Their product, \(6x^2\) is a perfect square. If \(x\) is a non-zero integer, then \(6\) is a perfect square, which is a contradiction.

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I think 0. If a = 2x and b = 3x then a = 2/3 * b So it would mean that root(2/3 * b) needs to be an integer. root(b) is an integer (cause it's a perfect square) but there is no integer that multiplied by an irrational number is also an integer. I'm not sure tho.

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no,1 is the answer.the only integer that can solve this is 0.

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Correct

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YEAH!! im thinking that also!!

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Copy paste this brute force code to jsbin.com function looker(){

var count=0;

for(i=0;i<100000000;i++){

if(perfect(2

i)===true && perfect(3i)===true){}

}

alert(count)

}

function perfect(N){

if(Math.floor(Math.sqrt(N))==Math.sqrt(N)){

}

else{return false

}

looker()

i checked up to 100000000 but only counts one

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