JBMO:NT

For discussing Olympiad Level Number Theory problems
aritra barua
Posts:57
Joined:Sun Dec 11, 2016 2:01 pm
JBMO:NT

Unread post by aritra barua » Tue Apr 25, 2017 4:10 pm

Consider the equation:$2^x$.$3^y$-$5^w$.$7^z$=$1$;$x$, $y$, $z$, $w$ are non negative integers.

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Atonu Roy Chowdhury
Posts:64
Joined:Fri Aug 05, 2016 7:57 pm
Location:Chittagong, Bangladesh

Re: JBMO:NT

Unread post by Atonu Roy Chowdhury » Wed Apr 26, 2017 12:13 am

It is obvious that $x \ge 1$ .
Case 1: $y=0$
Then, $2^x - 1 = 5^z . 7^w$
.
Subcase 1.1: $z > 0$
So, $2^x \equiv 1 (mod 5) $ . So, $ 4|x$. $2^{4k} - 1 = 16^k - 1$ which is divisible by $3$ . But $5^z . 7^w$ is not divisible by $3$.
.
Subcase 1.2: $z=0$
So, $2^x - 7^w = 1$
$x=1 \rightarrow w = 0$
$x=2 \rightarrow w$ is not integer
$x=3 \rightarrow w = 1$
$x \ge 4 \rightarrow 7^w \equiv -1 (mod 16) $ which is not possible.
.
Case 2: $y > 0$
.
Subcase 2.1: $x=1$
So, $2 . 3^y = 5^z . 7^w + 1$.
$5^z . 7^w \equiv -1 (mod 3)$ implies $z$ is odd. Again, $2.3^y \equiv 1 (mod 5) $ implies $y \equiv 1 (mod 4)$.
Now, if $w>0$, $2.3^y \equiv 1 (mod 7)$ implies $y \equiv 4 (mod 6)$. These two modular equation implies $y$ has no solution.
So, $w=0 \rightarrow 2.3^y - 5^z = 1$
$y=1 \rightarrow z=1$
$y \ge 2 \rightarrow 5^z \equiv -1 (mod 9) $ implies $3|z$ and so $5^3 + 1 | 5^z +1$. But $7 | 5^3 +1$
.
Subcase 2.2: $x \ge 2$
So, $5^z . 7^w \equiv -1 (mod 4)$ and $5^z . 7^w \equiv -1 (mod 3)$ implies $w$ and $z$ are odd.
$2^x . 3^y \equiv 5^z . 7^w +1 \equiv 36 (mod 8)$ implies $x=2$ . Then, $4 . 3^y \equiv 1 (mod 5)$ and $4 . 3^y \equiv 1 (mod 7)$ which gives $y \equiv 2 (mod 4)$ ans $y \equiv 2 (mod 6)$ . So, $y \equiv 2 (mod 12)$
$5^z . 7^w = (2 . 3^{6n+1} + 1) ( 2 . 3^{6n+1} - 1 )$
It can easily be found that
$2 . 3^{6n+1} + 1 = 7^w$ and
$ 2 . 3^{6n+1} - 1 = 5^z$
If $n \ge 1$ , $ 5^z \equiv -1 (mod 9) $ which is impossible.
$n=0 \rightarrow y = 2$
.
So, all the solutions are $(x,y,z,w) = (1,0,0,0),(3,0,0,1),(1,1,1,0),(2,2,1,1)$
This was freedom. Losing all hope was freedom.

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