RUSSIAN MATHEMATICAL OLYMPIAD

[MATHPRITOM]

Find all integer solutions of the equation $ (x^2-y^2)^2=1+16y $ .

[Phlembac Adib Hasan]

Beautiful equation.
My Solve :

The equation says\[(x^2-y^2+1)(x^2-y^2-1)=16y\]
\[\Rightarrow x^2\equiv 1,-1(mod\; y)\]
So we have two cases:
Case 1 :$x^2\equiv 1(mod\; y)$
Case 2 :$x^2\equiv -1(mod\; y)$

Case 1 :
Here $(x^2-y^2-1)$ contains $y$.So $(x^2-y^2+1)$ will be only a power of two.
Again I have four cases:\[x^2-y^2+1=1,2,4,8,16\]\[\Rightarrow x^2-y^2=0,1,3,7,15\]\[\Rightarrow (x,y)=(0,0),(1,0),(2,1),(4,3),(8,7)\]Checking these we find that only $(1,0)$ and $(4,3)$ works.Similarly from case 2, we find $(x,y)=(4,5)$.So all the solutions are \[(x,y)=(1,0),(4,3),(4,5)\]

[Masum]

The equation says\[(x^2-y^2+1)(x^2-y^2-1)=16y\]
\[\Rightarrow x^2\equiv 1,-1(mod\; y)\]
So we have two cases:
Case 1 :$x^2\equiv 1(mod\; y)$
Case 2 :$x^2\equiv -1(mod\; y)$

You said this after \[x^4\equiv1\pmod y\]
But unfortunately you can't conclude this. You can say about two cases: if $y=ab$ with $\gcd(a,b)=1$ then $a|x^2-y^2-1$ and $b|x^2-y^2+1$. Otherwise if $y=4ab$ and $2a|x^2-y^2-1$ or $2b|x^2-y^2+1$.
This is a common mistake. Someone did the same earlier.
Even if $a|bc$ with $\gcd(a,b)=1$ we can't say anything like $a|b$ or $a|c$ unless $a$ has only prime in its prime factorization.

[Phlembac Adib Hasan]

You said this after \[x^4\equiv1\pmod y\]
But unfortunately you can't conclude this. You can say about two cases: if $y=ab$ with $\gcd(a,b)=1$ then $a|x^2-y^2-1$ and $b|x^2-y^2+1$. Otherwise if $y=4ab$ and $2a|x^2-y^2-1$ or $2b|x^2-y^2+1$.
This is a common mistake. Someone did the same earlier.
Even if $a|bc$ with $\gcd(a,b)=1$ we can't say anything like $a|b$ or $a|c$ unless $a$ has only prime in its prime factorization.

Yes, you are right.It's a common mistake,but not here.I was not mistaken and did not write it unmindfully.Notice that two factors of LHS differ by $2$.So their gcd will be at most $2$.Hence I can conclude.

[Masum]

No. You haven't proven that in this case $y$ must divide one of them, think carefully. What you get is: \[a|x^2-y^2+1\Rightarrow a|x^2+1\]
Likewise, $b|x^2-1$. Then $ab|x^4-1$. Still not that \[y|x^2\pm1\]
\[(x^2-y^2+1)(x^2-y^2-1)=16y \]

\[x^2\equiv\pm1\pmod y \]
That was wrong. We have $x^4\equiv1\pmod y$. Not $x^2\equiv\pm1\pmod y$. For example $y=15,x=2$. Then \[y|x^4-1\]
But $y$ does not divide $x^2+1$ or $x^2-1$.

[Phlembac Adib Hasan]

Ok, if I think it in the following way, is it still mistake?
Let $y=k.2^m$ where $k$ is odd.As $x^2-y^2+1$ and $x^2-y^2-1$ have highest gcd $2$,it implies $k$ divides either $x^2-y^2+1$ or $x^2-y^2-1$.So $x^2\equiv 1,-1(mod \; k)$.The next part is similar to that of my previous proof.Here we find the same solutions as in before.Now is it still incorrect?

[Masum]

in fact, it is still incorrect. The example above shows this.
No matter. Try another approach. For example if $ab=x^2$ with $\gcd(a,b)=1$, then $a$ and $b$ are perfect squares.