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# perfect squre

no of positive integers of x such that 2x and 3x are to be a perfcct square

Note by Sai Venkata Raju Nanduri
4 years, 3 months ago

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Live question? · 4 years, 3 months ago

last week algebra level 3 · 4 years, 3 months ago

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. · 4 years, 3 months ago

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. · 4 years, 3 months ago

no,1 is the answer.the only integer that can solve this is 0. · 4 years, 3 months ago

Correct · 4 years, 3 months ago

With other questions I have found that they don't count 0 as a positive integer. · 4 years, 3 months ago

Read his comment, he said integer not positive integer, and 0 is an integer. · 4 years, 3 months ago

But the Question wasn't. · 4 years, 3 months ago

the answer for the problem was 1 · 4 years, 3 months ago

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. · 4 years, 3 months ago

Then I'm not the one making it difficult here. You could have just told that instead. Indeed just leave it. · 4 years, 3 months ago

YEAH!! im thinking that also!! · 4 years, 3 months ago

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

var count=0;

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

if(perfect(2i)===true && perfect(3i)===true){

count++


}

}

}

function perfect(N){

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

return true


}

else{return false

  }


}

looker()

i checked up to 100000000 but only counts one · 4 years, 3 months ago