Huan's Perfect Squares

Let's take A and B as two integers so that

${ A }^{ 2 }\quad =\quad 2x\\ { B }^{ 2 }\quad =\quad 3x$

Now, looking at this equation, we can clearly state that there is one solution, wich is

$x\quad =\quad 0$

Now, in order to look for other solutions : $A\quad =\quad \sqrt { 2x } \\ B\quad =\quad \sqrt { 3x }$

wich leads to :

$\frac { A }{ B } \quad =\quad \sqrt { \frac { 2 }{ 3 } }$

and since the ratio of two integers can't be an irrational number, we can clearly see that 0 is the only solution for the problem.

Let $A^2 = 2x$ and $B^2 = 3x$ . If we multiply them together then we have $(AB)^2 = 6x^2$ . Therefore $6x^2$ is a square number too but when $x\geq 1$ , it can't exist.

This can be shown by writing $x$ as it's prime factorisation like $x = 2^{p_1} \times 3^{p_2} \times 5^{p_3} \times \dots$ where $p_1, p_2, p_3, \dots$ , are positive integers. This means $6x^2 = 2^{2p_1 + 1} \times 3^{2p_2 + 1} \times 5^{2p_2} \times \dots$ . The problem now arises that for this to be a square number, all of the indices have to be even but the indices in 2 and 3 are odd. Therefore $x$ doesn't exist for $x\geq1$ .

It also doesn't work for $x \leq -1$ because $2x$ and $3x$ won't be positive and therefore not a perfect square. That only leaves $x = 0$ which works.

1 i think. I couldnt think of any number which i could multiply to 2 and 3 and would be a square. Just 0.

For all $n\in N$ , $n\neq 0$ we have a prime factorization ${ n }={ p }_{ 1 }^{ { \alpha }_{ 1 } }{ p }_{ 2 }^{ { \alpha }_{ 2 } }{ p }_{ 3 }^{ { \alpha }_{ 3 } }...{ p }_{ k }^{ { \alpha }_{ k } }$ . Is true that ${ { n }^{ 2 } }={ p }_{ 1 }^{ { 2\alpha }_{ 1 } }{ p }_{ 2 }^{ { 2\alpha }_{ 2 } }{ p }_{ 3 }^{ { 2\alpha }_{ 3 } }...{ p }_{ k }^{ { 2\alpha }_{ k } }$ . In the prime factorization of a perfect square we have all the exponents pair. If 2x is a perfect square than ${ 2 }^{ 2h+1 }|x$ . If 3x is a perfect square than ${ 3 }^{ 3t+1 }|x$ so $x={ 2 }^{ 2h+1 }\cdot { 3 }^{ 2t+1 }$ (because x musts contains only 2 and 3 prime factors) and we have

$\begin{cases} 2x={ 2 }^{ 2h+1 }\cdot { 3 }^{ 2t+1 } \\ 3x={ 2 }^{ 2h+1 }\cdot { 3 }^{ 2t+1 } \end{cases}\rightarrow \begin{cases} x={ 2 }^{ 2h }\cdot { 3 }^{ 2t+1 } \\ x={ 2 }^{ 2h+1 }\cdot { 3 }^{ 2t } \end{cases}\rightarrow { 2 }^{ 2h+1 }\cdot { 3 }^{ 2t }={ 2 }^{ 2h }\cdot { 3 }^{ 2t+1 }\rightarrow 2=3$ . That is absurd so x could be only 0: $2\cdot 0=3\cdot 0={ 0 }^{ 2 }$ .

There isn't any number that can be multiplied to 2 and 3 at the same time and be a perfect square except the number ZERO ( 0 ) . So the answer is ONE ( 1 )

2x=u^2,3x=b^2. (u^2)/2=(b^2)/3.u^2=2(b^2)/3. setting u = b =0 we have a trivial solution. Suppose there existed another solution. We know that b^2 is a multiple of 3. hence b has the prime factor 3 an even number of times. 2(b^2)/3 has the prime factor 3 an odd number of times hence it is not a perfect square which is a contradiction

There is one solution: x = 0.

Concept: for this to work, both 2x and 3x have to be perfect squares. Now, redefine 2x as "y." Then 3x is (3/2) y. But this means both have a rational square root; respectively, sqrt(y) and sqrt(3y/2). But this would give sqrt(3)/2 times sqrt(y); even if sqrt(y) is an integer, sqrt(3)/2 times that integer would remain irrational. Ergo, y and (3/2)y cannot both be perfect squares.