Equation with integers $x$, $y$

Question

If $x$, $y$ positive integers ($x<y$), how can I solve the equation $x+y=14\sqrt{xy-48}$ ?

Answer 1

Squaring yields the Diophantine equation $$ x^2 - 194xy + y^2 + 9408=0. $$ This type of Diophantine equation can be solved in general, see for example here, or here. Possible solution are $(x,y)=(1,97),(2,26), (26,2),(26,5042), (97,1),(97,18817), \cdots $. It reduces to a Diophantine equation of Pell type.

Answer 2

We have: $$x + y = 14\sqrt{xy -48} \implies 196(xy-48)= (x+y)^2 \implies x^2+y^2-194xy+9408 =0$$

Now, rearranging the terms we get: $$ y^2-194xy = -x^2-9408 \implies 9409x^2-194xy +y^2=9408x^2-9408$$ $$ \implies (y-97x)^2 = 9408(x^2-1) \implies y = 97x \pm 56\sqrt{3(x^2-1)}$$

There are many solutions to this: $(1,97), (2,26)$, etc.

Answer 3

Write $x^2 - 194xy + y^2$ in matrix form: $$ \pmatrix{x & y} \pmatrix{ 1 & -97 \\ -97 & 1} \pmatrix{x \\ y} $$ This matrix has eigenvectors $(1,\pm 1)$, which leads us to consider $u=x+y$ and $v=x-y$.

Then $x^2 - 194xy + y^2 + 9408=0$ becomes $49 v^2 - 48 u^2+ 9408=0$.

Since $9408 =2^6×3×7^2$, this implies that $u=7u_1$ and $v=12v_1$ and so $3v_1^2-u_1^2+4=0$, a Pell equation.

Answer 4

Say $z=x+y\geq 0$ then $$ z^2 = 14^2(zx-x^2)-14^2\cdot 48$$ so $z= 14t$ and thus we have $$t^2= 14tx-x^2-48$$ or $$t^2-14tx +49x^2 =48(x^2-1)$$ so $$(t-7x)^2= 48(x^2-1)$$ and now we have $t-7x = 12s$ for some integer $s$. So we have $$ 3s^2= x^2-1$$ If $3|x-1$ then $$x-1 =3u^2\;\;\;{\rm and} \;\;\;x+1 = v^2$$ where $u,v$ are relatively prime. In this case take mod 3 and we get $v^2 =_3 2$ which is impossible.

If $3|x+1$ then $$x+1 =3u^2\;\;\;{\rm and} \;\;\;x-1 = v^2$$ where $u,v$ are relatively prime. Then we get Pell's equation $3u^2-v^2=2$ which has infinitely solutions...