can you explain it more specificly

shinichiman wrote:

don't get the hint..

henderson wrote: Set $A$ consists of natural numbers such that these numbers can be expressed as $2x^2+3y^2,$ where $x$ and $y$ are integers. $(x^2+y^2\not=0)$ $a)$ Prove that there is no perfect square in the set $A.$ $b)$ Prove that multiple of odd number of elements of the set $A$ cannot be a perfect square. (a) For the sake of contradiction, assume that $2x^2+3y^2=n^2$ for some positive integer $n$. Consider the smallest solution $(x,y)$ such that $2x^2+3y^2$ is a square. We have the following cases. 1. $3|n$. Then $3|x$ as well. Assuming $n=3m$ and $x=3a$, \begin{align*} 6a^2+y^2 & = 3m^2 \end{align*}This again forces that $3|y$. Letting $y=3b$, we get $2a^2+3b^2 =m^2$. However, $(a,b)$ is a smaller solution than $(x,y)$, contradiction. 2. $3\nmid n$. Then we also have $3\nmid x$. So, $n^2\equiv1\pmod3$ and $x^2\equiv1\pmod3$. \begin{align*} 2x^2+3y^2 & \equiv1\pmod3\\ 2x^2 & \equiv1\pmod3\\ 2 & \equiv1\pmod3 \end{align*}A contradiction. 3. $2|n$. Then $2|y$ and in a similar fashion as above we reach a contradiction. 4. $2\nmid n$. So, $n^2\equiv y^2\equiv1\pmod8$. \begin{align*} 2x^2+3y^2 & \equiv1\pmod8\\ 2x^2+3 & \equiv1\pmod8\\ 2x^2 & \equiv-2\pmod8\\ x^2 & \equiv-1\pmod4 \end{align*}We can divide the congruence by $2$ since $\gcd(2,4)=2$ by modular cancellation. Again, this is a contradiction since $x^2\equiv0,1\pmod4$ for integer $x$. (b) \begin{align*} (2a^2+3b^2)(2c^2+3d^2) & = 4a^2c^2+6a^2d^2+6b^2c^2+9b^2d^2\\ &= 4a^2c^2+12acbd+9b^2d^2+6(a^2d^2-2abcd+b^2c^2)\\ & = (2ac+3bd)^2+6(ad-bd)^2\\ & = u^2+6v^2\\ (2x^2+3y^2)(u^2+6v^2) & = 2x^2u^2+12x^2v^2+3u^2y^2+18y^2v^2\\ &=2(x^2u^2+9y^2v^2)+3(u^2y^2+4x^2v^2)\\ &=2(x^2u^2+6uvxy+9y^2v^2)+3(u^2y^2-4uvxy+4x^2v^2)\\ &=2(xu+3yv)^2+3(uy-2xv)^2 \end{align*}This is not a square by (a)

For part a) we use infinite descent. Let $(x_{1}, y_{1}) $ be the smallest pair belonging to A such that it is a perfect square. By smallest we mean such that $x_{1}+y_{1}$ is the least. Then mod 3 gives $3|x_{1}$ but then since 3 divides a square,so does 9 and we get that $3|y_{1}$ but then we get that the expression is $18x_{2}^2 +27y_{2}^2$ which is 9 times a perfect square. So $(x_{2}, y_{2}) $ is a smaller solution pair and we are done. Part b uses part 1 just some more work.