Prove that there is no pair $(a,b)$ of integers such that
$a^2=b^7+7$
Search found 47 matches
- Sat Jul 01, 2017 3:40 pm
- Forum: Number Theory
- Topic: no solution (a,b)
- Replies: 2
- Views: 6402
- Sat Jul 01, 2017 3:35 pm
- Forum: Number Theory
- Topic: Yet divisibility...
- Replies: 6
- Views: 8075
Yet divisibility...
Determine all ordered pairs $(a,b)$ of positive integers for which $\dfrac{b^3+1}{ab-1}$ is an integer.
- Sat Jul 01, 2017 3:31 pm
- Forum: Algebra
- Topic: a+b+c>=1/a+1/b+1/c
- Replies: 3
- Views: 9868
a+b+c>=1/a+1/b+1/c
Let $a,b,c$ be positive real numbers, such that: $a+b+c \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}.$
Prove that:
\[a+b+c \geq \frac{3}{a+b+c}+\frac{2}{abc}. \]
Prove that:
\[a+b+c \geq \frac{3}{a+b+c}+\frac{2}{abc}. \]
- Sat Jul 01, 2017 3:24 pm
- Forum: Algebra
- Topic: Very strange inequality..
- Replies: 1
- Views: 5971
Re: Very strange inequality..
Let $a$,$b$,$c$ be real positive numbers. Prove that
\[\left(\frac{a^3+abc}{b+c}\right)+\left(\frac{b^3+abc}{c+a}\right)+\left(\frac{c^3+abc}{a+b}\right)\ge a^2+b^2+c^2\]
\[\left(\frac{a^3+abc}{b+c}\right)+\left(\frac{b^3+abc}{c+a}\right)+\left(\frac{c^3+abc}{a+b}\right)\ge a^2+b^2+c^2\]
- Sat Jul 01, 2017 3:24 pm
- Forum: Algebra
- Topic: Inequality with a,b,c sides of a triangle
- Replies: 7
- Views: 13449
Inequality with a,b,c sides of a triangle
Let $a,b$ and $c$ be sides of a triangle. Prove that:
$\frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a}-\sqrt{b}}+\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\leq 3$
$\frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a}-\sqrt{b}}+\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\leq 3$
- Sat Jun 24, 2017 3:31 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: 2007 number 5 - divisibility
- Replies: 1
- Views: 7083
2007 number 5 - divisibility
Let $a$ and $b$ be positive integers. Show that if $4ab-1$ divides $(4a^2-1)^2$, then $a=b$.
- Sun Jun 18, 2017 10:50 pm
- Forum: Number Theory
- Topic: Divisibility... with x,y
- Replies: 1
- Views: 2286
Divisibility... with x,y
Determine all pairs of positive integers $(x,y)$ such that
$(7^x-3^y)|(x^4+y^2)$
$(7^x-3^y)|(x^4+y^2)$
- Sun Jun 18, 2017 10:48 pm
- Forum: Number Theory
- Topic: n^5+n^4
- Replies: 1
- Views: 2406
n^5+n^4
Determine all pairs of positive integers $(m,n)$ such that
$n^5+n^4=7^m-1$
$n^5+n^4=7^m-1$
- Sun Jun 18, 2017 10:43 pm
- Forum: Algebra
- Topic: Very strange inequality..
- Replies: 1
- Views: 5971
Very strange inequality..
Let $a$,$b$,$c$ be real positive numbers. Prove that
\[\left(\frac{a^3+abc}{b+c}\right)+\left(\frac{b^3+abc}{c+a}\right)+\left(\frac{c^3+abc}{a+b}\right)\ge a^2+b^2+c^2\]
[/quote]
\[\left(\frac{a^3+abc}{b+c}\right)+\left(\frac{b^3+abc}{c+a}\right)+\left(\frac{c^3+abc}{a+b}\right)\ge a^2+b^2+c^2\]
[/quote]
- Sun Jun 18, 2017 10:36 pm
- Forum: Algebra
- Topic: Good inequality..
- Replies: 2
- Views: 8425
Good inequality..
Let a, b, c be real positive numbers. Prove that
$$(1+a/b)(1+b/c)(1+c/a)>=2(1+(a+b+c)/(\sqrt[3]{abc}))$$
$$(1+a/b)(1+b/c)(1+c/a)>=2(1+(a+b+c)/(\sqrt[3]{abc}))$$