Show that $x^4-20200y^2=1$ has no solution in postive integers Show that $x^4-20200y^2=1$ has no solution in postive integers.
This topic is a question for the Chinese middle school students' mathematics competition today, so I think this problem has a simple solution.
 A: The Pell equation $z^2-20200y^2=1$ has the fundamental solution:
$$
x^2=z = 30729653461384352052499,\;
y = 216213087258427446135
$$
But $z$ is not a perfect square. All remaining solutions are $(z_m,y_m)$ with $z_m+y_m\sqrt{20200}=(z+y\sqrt{20200})^m$. Then we have to show that $z_m$ is not a perfect square for all $m$. I suppose that this is not the idea for the middle school, but at least one has a solution.
A: We have to solve
$$
x^4-20200y^2=1\tag 1
$$
Observe that $20200=6^2+142^2$. Then (1) becomes 
$$
x^4=(6y)^2+(142y)^2+1.\tag 2
$$
Clearly $x=2k+1$ is odd. Hence $x^2=8T+1$,$T=\frac{k(k+1)}{2}$. Hence $x^4=8T_1+1=16T(4T+1)+1\Rightarrow T_1=2T(4T+1)=\frac{4T(4T+1)}{2}$. In general holds the following 
THEOREM. 
If $x$ is odd then $x^2=8T+1$ iff $T$ is triangular number.
PROOF. 
Easy.
But then (since $6=2\cdot3$ and $142=2\cdot 71$), we have $x^4=4(3y)^2+4(71y)^2+1$. Hence $\frac{x^4-1}{4}=(3y)^2+(71y)^2$. Thus $\frac{8T_1+1-1}{4}=2T_1=5050y^2$ and we arrive to 
$$
T_1=2T(4T+1)=2525y^2.\tag 2
$$
Hence $4T^2+T=2\cdot2525y_1^2$, since $2|y\Rightarrow y=2y_1$. Hence $2|T\Rightarrow T=2T'\Rightarrow 8T'^2+T'=2525y_1^2$. But exists $y_2$ such that $y_1^2=8y_2+1$. Hence $2525 y_1^2\equiv 5(mod 8)$. Hence $T'\equiv 5(mod 8)$. Hence 
$$
T\equiv 2(mod 8).\tag 3
$$ 
Lastly from (2): $2T(4T+1)=2525y^2$ we have $8T^2+2T=2525y^2$ and since $y^2\equiv 1(mod 8)$ and $2525\equiv 5(mod 8)$, we get $2T=5(mod 8)$, which is contradiction to (3).
Hence equation (1) is imposible.
