Solve it for $t$ . $ t =\frac{(x^2+y^2)(x+\sqrt{x^2+y^2})}{y^2}$ . Part a . Notice that the numerator is an increasing function in $x$ and $x$ is an positive integer , so $x$ must be $1$ Now we want to find the minimum value of $f(y)=\frac{(1+y^2)(1+\sqrt{1+y^2})}{y^2}$ . $f(1)=2(1+\sqrt2)$ Assume that $y\geq2$ and Let $u^2=1+y^2\geq 5$. Then $ f(y)=u+1+\frac{1}{u-1}$ . It's easy to prove that $f(y)$ is increasing for all real numbers $\geq2$ . $f(2)=\frac{5}{4}(1+\sqrt5$ , and it's clearly $f(2)<f(1)$ . Therefore $t=\frac{5}{4}(1+\sqrt5)$. Part b . Let $d=\gcd(x,y)$ and $x=da$ , $y=db$ . $a^2+b^2$ must be a perfect square . That is $(a,b)=(2mn,m^2-n^2)$ or $(a,b)=(m^2-n^2,2nm)$ with $m\not\equiv n (\mod 2)$ and $\gcd(m,n)=1$ $t=d\frac{(m^2+n^2)^2}{(m-n)^2}$ or $t=d\frac{(m^2+n^2)^2}{2n^2}$ First case : $\gcd(m^2+n^2,m-n)=1\Rightarrow (m-n)^2|d\Rightarrow t=k(m^2+n^2)^2$ Second case : $\gcd(m^2+n^2,2n^2)=1\Rightarrow 2n^2|d\Rightarrow t=k(m^2+n^2)^2$ (In first case $k=d/(m-n)^2$ , similarly in second case ) . Therefore the least poisitive integer $t$ satisfies the problem condition is $t=(1)(1^2+2^2)^2=25$

For the second part of this problem, I don't think we need any of this hassle. Simply, we have, for $x=dx_1$, $y=dy_1$ with $d={\rm gcd}(x,y)$, $$ ty_1^2 = d(x_1^2+y_1^2)(x_1 + \sqrt{x_1^2+y_1^2}). $$and seek for a $t$, that would lead the existence of such a pair. Clearly, both hand sides are integer, hence, simply note, $x_1^2+y_1^2 \mid t$, as $(x_1^2+y_1^2,y_1^2)=1$, hence, we guarantee that, $t$ is at least greater than or equal to $x_1^2+y_1^2$, for the smallest coprime pair, for which $x_1^2+y_1^2$ is a perfect square. We do know, smallest such pair is $(4,3)$ (equivalently, $(3,4)$, just check), hence, smallest $t$ is 25.