Question

Solving $(2^k-3)b^2 - (2^{k+1}+1)ab - (2^k+2)a^2 = 0$

score 181 · Answer 1 · 2016-08-22 23:51:44

181

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Solving $(2^k-3)b^2 - (2^{k+1}+1)ab - (2^k+2)a^2 = 0$

Published on 22 Aug 2016 - 23:51

score 8.3k · Answer 2 · 2016-08-29 00:46:58

8.3k

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Math olympiad 1988 problem 6: canonical solution (2) without Vieta jumping

Published on 29 Aug 2016 - 0:46

score 1k · Answer 3 · 2016-09-01 21:18:10

1k

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Prove that, if a and b are positive integers and $ab\mid a^2 + b^2$ then $a=b$

Published on 01 Sep 2016 - 21:18

score 1.7k · Answer 4 · 2016-09-07 01:41:05

Let $a$ and $b$ be positive integers such that $ab + 1$ divides $a^2 + b^2$. Show that $(a^2 + b^2)/(ab + 1)$ is a perfect square.

score 1.8k · Answer 5 · 2016-09-17 16:01:45

1.8k

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Equation with Vieta Jumping: $(x+y+z)^2=nxyz$.

Published on 17 Sep 2016 - 16:01

score 2.5k · Answer 6 · 2016-10-31 12:24:54

2.5k

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Vieta Jumping: Related to IMO problem 6, 1988: If $ab + 1$ divides $a^2 + b^2$ then $ab + 1$ cannot be a perfect square.

Published on 31 Oct 2016 - 12:24

score 2.1k · Answer 7 · 2017-03-15 22:32:29

2.1k

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Showing that $m^2-n^2+1$ is a square

Published on 15 Mar 2017 - 22:32

score 419 · Answer 8 · 2017-03-26 23:35:07

419

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History of Vieta Jumping method

Published on 26 Mar 2017 - 23:35

score 96 · Answer 9 · 2023-08-06 19:31:45

96

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Is this valid? $r_1^3+r_2^3+\cdots+r_7^3=(r_1+\cdots + r_7)^3-3(r_1+\cdots+r_7)(r_1r_2+r_1r_3+\cdots+r_6r_7)+3(r_1r_2r_3+r_1r_2r_4+\cdots+r_5r_6r_7)$

Published on 06 Aug 2023 - 19:31

score 389 · Answer 10 · 2015-01-24 09:30:46

389

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Integer solutions to $\prod\limits_{i=1}^{n}x_i=\sum\limits_{i=1}^{n}x_i^2$

Published on 24 Jan 2015 - 9:30

score 266 · Answer 11 · 2015-01-26 08:38:43

266

Views

If $\frac{a+1}{b}+\frac{b}{a}$ is an integer then it is $3$.

Published on 26 Jan 2015 - 8:38

score 478 · Answer 12 · 2015-02-09 15:18:42

478

Views

Solve: $ab+bc+ca\mid (a+b+c)^2$

Published on 09 Feb 2015 - 15:18

score 1.3k · Answer 13 · 2015-04-28 09:16:21

1.3k

Views

Is it possible to generate a solution for this? $p^2 + q^2 + r^2 = pqr $

Published on 28 Apr 2015 - 9:16

score 415 · Answer 14 · 2015-06-18 17:52:59

415

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Integer solutions to the equation $a_1^2+\cdots +a_n^2=a_1\cdots a_n$

Published on 18 Jun 2015 - 17:52

score 303 · Answer 15 · 2015-08-17 06:52:05

303

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Determine all positive integers $n$ which have a divisor $d$ with the property that $dn+1$ is a divisor of $d^2 + n^2$

Published on 17 Aug 2015 - 6:52

List Question

Solving $(2^k-3)b^2 - (2^{k+1}+1)ab - (2^k+2)a^2 = 0$

Math olympiad 1988 problem 6: canonical solution (2) without Vieta jumping

Prove that, if a and b are positive integers and $ab\mid a^2 + b^2$ then $a=b$

Let $a$ and $b$ be positive integers such that $ab + 1$ divides $a^2 + b^2$. Show that $(a^2 + b^2)/(ab + 1)$ is a perfect square.

Equation with Vieta Jumping: $(x+y+z)^2=nxyz$.

Vieta Jumping: Related to IMO problem 6, 1988: If $ab + 1$ divides $a^2 + b^2$ then $ab + 1$ cannot be a perfect square.

Showing that $m^2-n^2+1$ is a square

History of Vieta Jumping method

Is this valid? $r_1^3+r_2^3+\cdots+r_7^3=(r_1+\cdots + r_7)^3-3(r_1+\cdots+r_7)(r_1r_2+r_1r_3+\cdots+r_6r_7)+3(r_1r_2r_3+r_1r_2r_4+\cdots+r_5r_6r_7)$

Integer solutions to $\prod\limits_{i=1}^{n}x_i=\sum\limits_{i=1}^{n}x_i^2$

If $\frac{a+1}{b}+\frac{b}{a}$ is an integer then it is $3$.

Solve: $ab+bc+ca\mid (a+b+c)^2$

Is it possible to generate a solution for this? $p^2 + q^2 + r^2 = pqr $

Integer solutions to the equation $a_1^2+\cdots +a_n^2=a_1\cdots a_n$

Determine all positive integers $n$ which have a divisor $d$ with the property that $dn+1$ is a divisor of $d^2 + n^2$

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