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.

Solution 1:

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.

Solution 2:

Continuing @JoshuaZ's comment:

Can make $z=10y$, and equation becomes $x^4−202z^2=1$. $x$ must be odd, so set $x=2k+1$. Then this is $(2k+1)^4−1=202z^2$ which is the same as $4k(k+1)(2k^2+2k+1)=101z^2$. Note that $k, k+1$ and $2k^2+2k+1$ are all relatively prime, so two are perfect squares and one is $101$ times a square. $k$ and $k+1$ cannot both be perfect squares, so one of them is a perfect square and one is $101\cdot \text{square}$, and then $2k^2+2k+1$ is a perfect square. So, if we can show this combination cannot happen then we are done.

... as $k^2+(k+1)^2$ is square, $k$ and $k+1$ are $u^2-v^2$ and $2uv$ in some order, for some $u,v\in\mathbb{N}$ coprime. Thus $u^2-v^2$ and $2uv$ are $s^2$ and $101t^2$ in some order, for some $s,t\in\mathbb{N}$.

Case 1: If $2uv=s^2$, then $u,v$ are $p^2$ and $2q^2$ in some order, for $p,q\in\mathbb{N}$ coprime and $p$ odd. If $u=2q^2$ and $v=p^2$ then $4q^4-p^4=101t^2$, which is impossible mod $4$.

If $u=p^2$ and $v=2q^2$ then $p^4-4q^4=101t^2$, so since $\gcd(p^2-2q^2,p^2+2q^2)=1$, $p^2-2q^2,p^2+2q^2$ are $c^2$ and $101d^2$ in some order. However the equation $p^4-4q^4=101t^2$ mod $8$ yields $q$ odd, so that $p^2-2q^2$ and $p^2+2q^2$ are $3$ mod $4$, impossible.

Case 2.1: If $2uv=101t^2$ and $u,v$ are $101p^2$ and $2q^2$ in some order, for $p,q\in\mathbb{N}$ coprime and $p$ odd, then if $u=2q^2$ and $v=101p^2$ then $4q^4-101^2p^4=s^2$, which is impossible mod $4$.

If $u=101p^2$ and $v=2q^2$ then $101^2p^4-4q^4=s^2$, so similarly to the last case, $101p^2-2q^2$ and $101p^2+2q^2$ are both squares. The equation $101^2p^4-4q^4=s^2$ mod $8$ yields $q$ even, so that $101p^2-2q^2$ and $101p^2+2q^2$ are both $5$ mod $8$, which is impossible.

Case 2.2: If $2uv=101t^2$ and $u,v$ are $p^2$ and $202q^2$ in some order, for $p,q\in\mathbb{N}$ coprime and $p$ odd, then if $u=202q^2$ and $v=p^2$ then $202^2q^4-p^4=s^2$, which is impossible mod $4$.

If $u=p^2$ and $v=202q^2$ then $p^4-202^2q^4=s^2$, so $p^2=a^2+b^2$ and $202q^2=2ab$ for some $a,b\in\mathbb{N}$ coprime and differ in parity. Then $a^2+b^2=u\le (u^2+v^2)^2=k^2+(k+1)^2$. Note that the only things we've used about $k$ and $k+1$ in this proof is that they are coprime and differ in parity. Therefore if we repeat the proof but with $a,b$ in place of $k,k+1$, we can generate a smaller set of $a,b$. By Fermat's infinite descent, the only solution is when $a^2+b^2=k^2+(k+1)^2$, which implies that $k=0$ and $(x,y)=(1,0)$. (It's easy to check it's the only solution that descends to itself.)

Solution 3:

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.

Solution 4:

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.