2^x * 3^y - 5^z * 7^w = 1

orl

Find all nonnegative integer solutions $(x,y,z,w)$ of the equation $2^x\cdot3^y-5^z\cdot7^w=1.$

**Posted:** Tue Jan 25, 2005 7:39 am

_________________
Math is like love. A simple idea but it can get complicated.

pengshi

$(x,y,z,w) = (1,1,1,0)$ , $(2,2,1,1,)$ , $(1,0,0,0)$ or $(3,0,0,1)$

If $x = 0$ , then we get odd-odd=odd, contradiction

If $x\geq 3$ , $y \geq 1$ , then mod 3, we get $z$ is odd, mod 8, we get $z$ is even, contradiction.

if $y \geq 1$ , $2\geq x \geq 1$ then [

if $x=1$ , then if $z\geq 1$ , $w \geq 1$ [
mod 7, we find that $y=4$ (mod 6), which means that $y$ is even. mod 5, we find that $y=1$ mod 4, which means that $y$ is odd, contradiction.

]else if $z = 0$ , $w \geq 0$ [ If $y=0$ , then we get the solution $(1,0,0,0)$ . we check that $y=1$ dosen't work, so mod 9, we get a contradiction.

]else if $z > 0$ , $w = 0$ [ If $y \leq 1$ , we check to find the solution $(1,1,1,0)$ . Otherwise, mod 9, and we find that $z$ is divisible by 3. However since $5^{3k} +1$ is divisible by 7, we get a contradiction.
]

if $x=2$ , then [ we find the solution $(2,2,1,1)$ . Mod 7 or mod 5, we find that y is even. Let $y= 2g$ . If y is larger than 4 then $5^z7^w = 4*3^y -1 = (2*3^g-1)(2*3^g+1)$ . We find that $(2*3^g-1)$ must equal to $5^z$ or $7^w$ , where $g>1$ . We have already proven that this is impossible. (this is equivalent to the previous two cases.)
]
]

Now, consider the case when $y=0$ . Mod 7, we find that $x$ is divisible by 3. If $z \geq 1$ , then mod 5, we find that $x$ is divisible by 4. However, since $2^{2k}-1$ is divisible by 3, thus, contradiction. Finally, we are left with the case that $y=0$ , $z=0$ . If $x\leq3$ , then we check and find the solution $(3,0,0,1)$ , and if $x>3$ , mod 16 would yield a contradiction.

Q.E.D.
P.S. There got to be a better solution!

**Posted:** Tue Jan 25, 2005 6:42 pm

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...and the rest follows by induction.

nhat

i have the other solution
first we easy to prove that x <= 2 then we have two case:
if x=1 then easu to prove that the equation hasn't root (use modular)
if x=2 then we mast use Pell's equation and we have the result (for the case two it's two long and i am very lazy to write it) Mr. Green

**Posted:** Wed Jan 26, 2005 3:55 am

ikap

maybe I'm blind, but how do you particularly reach pell's equation in here. And even if u do I don't think it gets easier than the ordinary solution.

**Posted:** Fri Feb 11, 2005 10:24 am

Erken

**Posted:** Sat Jun 14, 2008 4:10 am