## Search found 12 matches

Tue Sep 10, 2013 12:16 pm
Forum: Number Theory
Topic: $2^{x}=3^{y}+5$
Replies: 4
Views: 1632

### Re: $2^{x}=3^{y}+5$

asif e elahi wrote:
harrypham wrote: Therefore $3^y \equiv 59 \pmod{64}$. It follows that $y \equiv 11 \pmod{16}$. Hence $3^y \equiv 3^{11} \equiv 7 \pmod{17}$.
How $y \equiv 11 \pmod{16}$ ?
You note that $3^{16} \equiv 1 \pmod{64}$ and $3^{11} \equiv 59 \pmod{64}$.
Mon Sep 09, 2013 12:36 pm
Forum: National Math Camp
Topic: Equation!
Replies: 5
Views: 4782

### Re: Equation!

SANZEED wrote:Find all real solution of $\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$.
The equation is equivalent to $|\sqrt{x-1}-2|+|\sqrt{x-1}-3|=1$.
From here we use the inequality $|a|+|b| \ge |a+b|$.
Mon Sep 09, 2013 12:01 pm
Forum: Number Theory
Topic: $2^{x}=3^{y}+5$
Replies: 4
Views: 1632

### Re: $2^{x}=3^{y}+5$

More stronger: $2^x=3^y+5^z$.
Mon Sep 09, 2013 11:59 am
Forum: Number Theory
Topic: $2^{x}=3^{y}+5$
Replies: 4
Views: 1632

Find all positive integers x,y such that $2^{x}=3^{y}+5$ It is easy to see that $(x,y)=(3,1),(5,3)$. If $x \ge 6$ then $64|2^x$. Therefore $3^y \equiv 59 \pmod{64}$. It follows that $y \equiv 11 \pmod{16}$. Hence $3^y \equiv 3^{11} \equiv 7 \pmod{17}$. Since $3^y+5 \equiv 2 \pmod{3}$ then $2^x \equ... Sun Aug 11, 2013 9:06 pm Forum: Number Theory Topic: modulo Replies: 1 Views: 1018 ### Re: modulo We have$m|a-b$with$m,a,b \in \mathbb{Z}, \; m \ne 0$then we write$a \equiv b \pmod{m}$. Some property:$a \equiv b \pmod{m}, \; b \equiv c \pmod{m}$then$a \equiv c \pmod{m}$.$a \equiv c \pmod{m}, \; b \equiv d \pmod{m}$then$ab \equiv cd \pmod{m}$.$a \equiv b \pmod{m}$then$a+c \equiv b+c...
Sat Aug 10, 2013 7:35 am
Forum: Number Theory
Topic: Can you find an integer b such that:3b^+3b+7？
Replies: 2
Views: 1469

### Re: Can you find an integer b such that:3b^+3b+7？

liyuqingru wrote:Find $b$ such that $3b^2+3b+7\vdots b$
Who can help me solve this question in detail? Thanks a lot!
From here we obtain $b|7$.
Sat Aug 10, 2013 7:30 am
Forum: Number Theory
Topic: Pairs of integers
Replies: 2
Views: 1683

### Re: Pairs of integers

SANZEED wrote:Find infinitely many pairs of integers $(a,b)$ such that $1<a<b$ and $ab$ divides $a^{2}+b^{2}-1$. Also find all positive integers $k$ such that there exists $(a,b)$ such that $\frac {a^{2}+b^{2}-1}{ab}=k$.
We use Vieta Jumping to solve this problem. Sat Aug 10, 2013 7:26 am
Forum: Secondary: Solved
Topic: Dhaka Secondary 2010/12
Replies: 5
Views: 5589

### Re: Dhaka Secondary 2010/12

Sat Aug 10, 2013 7:21 am
Forum: Secondary Level
Topic: Solve it!
Replies: 5
Views: 2479

### Re: Solve it!

Phlembac Adib Hasan wrote:Find all non-negative integer $x,y$ such that $y^2+1=2^x$
(It's a mathlinks problem!I solved yesterday.)
If $x \ge 2$ then $4|y^2+1$. Therefore $y^2 \equiv 3 \pmod{4}$, a contradiction.
Thus, $x=1$ or $x=0$.
Thu Aug 08, 2013 12:06 pm
Forum: Secondary Level