Find all non-negative integer solutions of the equation \[2^x + 3^y = z^2 .\]
Problem
Source: Iran Third Round 1996, E2, P1
Tags: number theory proposed, number theory
sumanguha
25.03.2011 15:51
Let $x\geq 2$ then considering modulo 4 we get $y=2k$ from there we get $ 2^x=(z-3^k)(z+3^k)=2^u2^v$ see $ 2^u>2^v$ so, $2.3^k=2^u(2^{v-u}-1)$ so, $u=1$ so, $z-3^k=2$ so, $z+3^k=2(1+3^k)=2^v$ so, $2^{v-1}-3^k=1$ now, considering modulo 3 $2^{2t}-3^k=1$ or, $3^k=(2^t-1)(2^t+1)=3^{k_{1}}3^{k_{2}}$ so, $3^{k_{1}}+2=3^{k_{2}}$ so, $k_{1}=0,k_{2}=1$ so, $k=1$ so, $z=5$ so, $(x,y,z)=(4,2,5)$ if $x=1$ then $2+3^y=z^2$ considering modulo 3 we see only possibility is $y=0$ which don't give a solution. $x=0$ then $3^y=(z-1)(z+1)$ gives $z-1=1$ so, another solution $(x,y,z)=(0,1,2)$ so, only solutions are $(x,y,z)=(0,1,2);(4,2,5)$
SKhan
17.08.2011 21:01
When $y=0$, we have $2^x+1=z^2\implies 2^x=(z-1)(z+1)$
Now $z+1>0$ and $2^x>0 \implies z-1>0$.
So $z-1$ and $z+1$ are positive integer factors of $2^x$, thus $z-1$ and $z+1$ are powers of $2$.
$z+1=2+(z-1)>2$ so $4|z+1$, so $z-1\equiv 2 \mod 4\implies z-1=2$
Then $z=3$ and $2^x=8\implies x=3$
This gives the solution $(x, y, z) = (3, 0, 3)$
Otherwise, $y>0\implies 3^y\equiv 0 \mod 3\implies z^2\equiv 2^x \mod 3$
Now $2^x$ is coprime to $3$ so $3\not |z^2\implies z^2\equiv 1 \mod 3$
So $2^x\equiv z^2\equiv 1 \mod 3\implies (-1)^x\equiv 1 \mod 3\implies x$ is even.
Let $x=2m$ where $m$ is a non-negative integer.
Then $3^y=(z-2^m)(z+2^m)$
$z+2^m>0$ and $3^m>0\implies z-2^m>0$.
So $z-2^m$ and $z+2^m$ are positive factors of $3^y$, thus $z-2^m$ and $z+2^m$ are powers of $3$.
$z+2^m>1$ so $3|z+2^m$.
$3\not |2^{m+1}\implies 3\not |z-2^m\implies z=2^m+1$.
Then $3^y= 2^{m+1}+1$
When $m=0$, we have $x=0$, $z=2^0+1=2$ and $3^y=2+1=3\implies y=1$
This gives the solution $(x, y, z) = (0, 1, 2)$
Otherwise, $m>0\implies 2^{m+1}\equiv 0\mod 4\implies 3^y=2^{m+1}+1\equiv 1\mod 4$
Then $(-1)^y\equiv 1\mod 4\implies y$ is even.
Let $y=2n$ where $n$ is a positive integer.
Then $(3^n-1)(3^n+1)=2^{m+1}$
$3^n+1>0$ and $3^n-1>0\implies 3^n+1$ and $3^n-1$ are positive integer factors of $2^{m+1}$, and hence are powers of $2$.
Now $3^n+1>2 \implies 4|3^n+1\implies 3^n-1\equiv 2 \mod 4\implies 3^n-1=2$
Then $n=1\implies y=2\implies 2^{m+1}=3^2-1=8\implies m=2\implies x=4$ and $z=2^2+1=5$
This gives the solution $(x, y, z) = (4, 2, 5)$
So all solutions are $(x, y, z) = (0, 1, 2); (3, 0, 3); (4, 2, 5)$
sayantanchakraborty
10.05.2014 09:06
Checking modulo 3 and modulo 4 we get x and y as even.Let $x \ge 2,y \ge 1$ and set $x=2k,y=2l$.Then the equation is equivalent to $2^{2k}=(z-3^l)(z+3^l)$.Hence $z+3^l=2^{2k-i},z-3^l=2^i$ for some nonnegative i.Subtracting these we get $2*3^l=2^i(2^{2k-2i}-1) \Rightarrow i=1$Thus $3^l=2^{2k-2}-1$.Now $l>1$ forces a contradiction modulo 8,since LHS mod 8 is 3 or 1 while the RHS mod 8 is 7.Thus it suffices to check for $l=1$ for which we get $k=2$.Thus $(x,y,z)=(4,2,5)$. As far as I remember this problem also appeared in the INMO some year(probably 1992).
spikerboy
10.05.2014 09:50
Amir Hossein wrote: Find all non-negative integer solutions of the equation \[2^x + 3^y = z^2 .\] If $y$ is positive then after taking $\mod 3$ we get, $x$ is even.Let, $x=2x_1$. $3^y=z^2-2^{2x_1} \implies 3^y=(z+2^{x_1})(z-2^{x_1})$.Since $z$ is odd.$\gcd(z+2^{x_1},z-2^{x_1})=1$ so,$z+2^{x_1}=3^a$ and $z-2^{x_1}=1$ for some $a \ge 0$ .so $3^{a} -1=2^{(x_{1}+1)}$.If $x_1 \ge 3 $ for $a \ge 3$ by zsigmondy's theorem $3^a -1$ has a prime factor $p$ that does not divide $3^2-1=8$ so there is no solution in this case.and we get by insepection $a=2, x_1=2$ and $a=1,x_1=0$ are the solutions.corresponding solutions are $(x,y,z)=(4,2,5),(0,1,2)$ and if $y$ is $0$ we get the solution $(x,y,z)=(3,0,3)$.So the total solutions are $(x,y,z)=(4,2,5),(0,1,2),(3,0,3)$
sayantanchakraborty
10.05.2014 10:48
Oh yes...I forgot the cases when x or y is 0.Sorry for that.
MathLuis
23.03.2022 03:40
Case 1: $x=0$ Then we have $z^2-3^y=1$ and if $y>1$ then we get a contraciction by Mihailescu theorem. Hence $y=0,1$ but if $y=0$ then $z^2=2$ which is not possible hence $y=1$ which means that $z=2$ so the pair $(x,y,z)=(0,1,2)$ works. Case 2: $x>0$ Using mod 3 we get thar $x$ is even so $x \ge 2$ and now by mod 4 we get $y$ even. Clearly $z$ is odd and now let $x=2a$ and $y=2b$ then, first if $b=0$ then $z^2-2^x=1$ and by Mihailescu we have $z=3$ and $x=3$ so the pair $(x,y,z)=(3,0,3)$ works. If $y \ge 2$ then $3^{2b}=(z+2^a)(z-2^a)$ and since $\gcd(z+2^a,z-2^a)=\gcd(2^{a+1},z-2^a)=1$ then $z+2^a=3^k$ and $z-2^a=3^l$ so now if both $k,l$ were positive integers then $2^{a+1}$ would be a multiple of 3 which is not possible so $l=0$ as $k>l$, now replace $z=2^a+1$ on the first equation to get $3^{2b}-2^{a+1}=1$ and by Mihailescu we get $b=1$ and $a=2$ so the pair $(x,y,z)=(4,2,5)$ works. Hence all the pairs that work are $\boxed{(x,y,z)=(0,1,2), (3,0,3), (4,2,5)}$ thus we are done