TITLE: Show that $x^4-20200y^2=1$ has no solution in postive integers
QUESTION [15 upvotes]: 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.

REPLY [2 votes]: We start with a simple well known lemma.
Lemma 1. $n^2 \equiv 1\, mod(8)$ for any odd number $n$.
And a well-known theorem.
Theorem 1. The triple $(x, y, z)$ is a primitive Pythagorean triple ($x^2+y^2=z^2$ and $x, y$ are coprime) if and only if there exist two relative prime integers $s$ and $t$ so that $s > t > 0$ and $x = 2st, y = s^2 − t^2$ and $z = s^2 + t^2$
where one of $s$ and $t$ is even and the other is odd.
Now we turn attention to our problem at hand.
Claim 1. $x$ is odd, and $y$ is even. Let $y=2^ez, e \ge 1$, where $z$ is an odd number.
Proof: Since $x$ is obviously odd, $8 \mid x^2 - 1$ due to Lemma 1. Let $x^2=8k+1$, one can easily see $16 \mid x^4 - 1$, or $16 \mid 20200y^2 \Rightarrow 2 \mid y.$ Hence $y$ is even. 
We rewrite the equation as (LHS & RHS that follow refer to this equation)
$$ (x-1)(x+1)(x^2+1)=101 \times 5^2 \times 2^{3+2e} \times z^2. $$
Claim 2. For any prime factor $p \ne 2$ of the RHS. Only one of the three terms of the LHS can contain factors of $p$, i.e. $p$ is not shared among the terms.
Proof. If $p \mid x-1$ and $p \mid x+ 1$, then $p \mid 2$, a contradiction.
If $p \mid x-1$ and $p \mid x^2+1$, then $p \mid x(x+1) \Rightarrow p \mid (x+1)$, a contradiction since we just showed $p \mid x-1$ and $p \mid x+ 1$ is not possible. If $p \mid x+1$ and $p \mid x^2+1$, then $p \mid x(x-1) \Rightarrow p \mid (x-1)$, again a contradiction.
For every prime factor $p \notin \{2, 101\}$ of RHS, the RHS ALWAYS has even number of factors of $p$, so is the LHS. By Claim 2, all the three terms of LHS are square numbers when factors of 2 and 101 are stripped away, and these square numbers are coprime.
We next consider the distribution of the prime factor of $2$. Note $x^2+1$ has exactly one $2$ (since $x^2$ + 1 = 8k+2). One of two items, $x-1$ and $x+1$, has exactly one $2$, and the other one gets all the remaining factors of 2, i.e $2^{1+2e}$.
Case 1. $101 \mid x^2+1$.
We have $x-1=2^{1+2e}s^2=2(2^es)^2$ and $x+1=2t^2$, or $x-1=2s^2$ and $x+1=2(2^et)^2$. Consolidate the two cases, we can write $x-1=2u^2, x+1=2v^2$. This is not possible since subtracting the two equations lead to $(v-u)(v+u)=1$ ($v=1, u=0$ is not a valid solution). 
Case 2 $101 \nmid x^2 + 1$.
Then we can write $x^2 + 1=2u^2$, $u$ is odd ($x^2 + 1$ does not contain 101, has only one 2, and all other prime factors are squared).
Note we already stated earlier that either $x-1$ or $x+1$ contains $2^{1+2e}$. And one of the two terms has factor 101.
Case 2.1 $101 \mid x+1$ and $2^{1+2e} \mid x+1$.
Then $x-1=2m^2$ for some odd number $m$, or $x=2m^2+1$. Then $$u^2=\frac{x^2+1}{2}=2m^4+2m^2+1$$.
Using lemma 1 and take modulus 8 over both sides leads to $1 \equiv 2+2+1\equiv 5$ mod (8), a contradiction. 
Case 2.2 $101 \mid x+1$ and $2^{1+2e} \mid x-1$.
Then $x + 1=2\times 101 \times m^2$ and $x - 1=2^{1+2e} n^2$ for some odd $m$ and $n$. Then $101 \times m^2 = 2^{2e} n^2 + 1$. Since $101 \times m^2 \equiv 5 mod(8),$ we have $e=1$ and $x - 1=8n^2$. Using $2u^2=x^2+1=\frac{{(x+1)}^2 + {(x-1)^2}}{2}$, we get ${\left(101m^2\right)}^2 + {\left(4n^2\right)}^2 = u^2$. 
Applying theorem 1 about the Pythagorean triples, there must be some coprime $s, t$ so that $101m^2=s^2-t^2$ and $4n^2=2st$ (note $101m^2 \ne 2st$), or equivalently $2n^2=st$. 
$s$ can not be even since otherwise $1  \equiv 101m^2  = s^2  - t^2 \equiv -1(mod 4).$ Hence $t$ must be even. Furthermore, since $s, t$ are coprime and $2n^2=st$, $s, t/2$ are square numbers. Let $t=2{t^\prime}^2, s={s^\prime}^2$, $s^\prime, t^\prime$ are odd and coprime. Then $101m^2={s^\prime}^4-4{t^\prime}^4=({s^\prime}^2-2{t^\prime}^2)({s^\prime}^2+2{t^\prime}^2)$. 
Further note $({s^\prime}^2-2{t^\prime}^2)$ and $({s^\prime}^2+2{t^\prime}^2)$ are coprime, one of them must be $101{m^\prime}^2$, $m^\prime \mid m$. ${s^\prime}^2-2{t^\prime}^2 = 101{m^\prime}^2$ is invalid since taking modulus 8 on both sides leads to $-1 \equiv 5 mod(8)$. ${s^\prime}^2+2{t^\prime}^2 = 101{m^\prime}^2$ is invalid as well since it leads to $3 \equiv 5 mod(8)$. Hence there is a contradiction.
We've shown this case provides no solution.
Case 2.3 $101 \mid x-1$ and $2^{1+2e} \mid x-1$.
Then $x = 2n^2-1$, $x = 2^{1+2e}\times 101 \times m^2+1$ for some odd number $m,n$. 
Similar to the previous step, we have 
$$u^2={(n^2)}^2 + {\left(2^{2e}\times 101 \times m^2\right)}^2\quad\text{  (1)}$$. 
Application of theorem 1 leads to $n^2 = s^2 - t^2$ and $2^{2e-1}\times 101 \times m^2 = st$. Again $s$ can not be even (otherwise, $n^2 = s^2 - t^2$ leads to $1 \equiv -1 mod (4)$). Hence $s$ is odd, $t$ is even. Prime number 101 can be assigned to both $s, t$ for the moment. Apart from prime factors of 2 and 101, $s$ and $t$ must be square numbers since they are coprime and their product is a square excluding 2 and 101 factors. 
The two possibilties are $s=101\times {s^\prime}^2, t=2^{2e-1}{t^\prime}^2$, or $s={s^\prime}^2, t=101\times 2^{2e-1}{t^\prime}^2, s^\prime t^\prime = m$.
If $s=101\times {s^\prime}^2, t=2^{2e-1}{t^\prime}^2$, then $n^2 = {\left(101\times {s^\prime}^2\right)}^2 - {\left(2^{2e-1}{t^\prime}^2\right)}^2 = \left(101\times {s^\prime}^2 - 2^{2e-1}{t^\prime}^2 \right)\left(101\times {s^\prime}^2 + 2^{2e-1}{t^\prime}^2\right).$ Notice the two terms on the right hand side are coprime, each of them must be a square number. Hence $101\times {s^\prime}^2 - 2^{2e-1}{t^\prime}^2 = {n^\prime}^2$. This is a contradiction since take modulus 8 leads to $5-2=3\equiv 1(mod 8)$ if $e=1$, and $5 \equiv 1(mod 8)$ if $e \ge 2$.
If $s={s^\prime}^2, t=101\times 2^{2e-1}{t^\prime}^2$, then $n^2={\left({s^\prime}^2\right)}^2 - {\left(101\times 2^{2e-1}{t^\prime}^2\right)}^2$. Because the factors 101 and 2 are sticking together, we could not use the trick in the last paragraph to get a contradiction. We can keep on applying theorem 1, i.e. $101\times 2^{2e-2}{t^\prime}^2=uv, {s^\prime}^2=u^2 + v^2.$ If 101 & 2 are separated into different components ($u$ and $v$), then we can apply tricks from the last paragraph to find a contradiction. So we continue with the assumption that they would be kept together, i.e. $u = 101\times 2^{2e-2}{u^\prime}^2, v={v^\prime}^2.$ Then 
$${s^\prime}^2 = {\left({v^\prime}^2\right)}^2 + {\left(101\times 2^{2e-2}{u^\prime}^2\right)}^2.\quad \text{  (2)}$$
Comparing equation (2) and (1), it is evident the difference, besides different names used for symbols of the same nature, is that the power of 2 is reduced from $2e$ to $2e-2$. We can repeat the argument process above iteratively until the power of 2 becomes zero in which case the equation won't hold, and that would be the contradiction.
Case 2.4 $101 \mid x-1$ and $2^{1+2e} \mid x+1$.
Then $x = 2\times 101 \times m^2+1$ for some odd number $m$. Then 
$$u^2=\frac{x^2+1}{2}=2\times 101^2 \times m^4+2\times 101 \times m^2+1$$.
Using Lemma 1 and take both sides modulus 8, the left side is 1 mod (8) while the right hand side is $2 + 2 * 5 + 1\equiv 5$ mod (8), a contradiction.
Now the proof is complete.