Determine all solutions of the diophantine equation $a^2 = b \cdot (b + 7)$ in integers $a\ge 0$ and $b \ge 0$. (W. Janous, Innsbruck)
Problem
Source: Austria Beginners' Competition 2014 p1
Tags: number theory, Diophantine equation, diophantine
Math-wiz
25.02.2020 18:33
We look it as a quadratic in $b$
$b=\frac{-7+\sqrt{49+4a^2}}{2}$
So, $49+4a^2=m^2$ for some $m\in\mathbb{Z}$
$49=(m-2a)(m+2a)$
Now just case bash
Kazeki_Kuroko
25.02.2020 18:43
4a^2=4b^2+28b+49 - 49 =>(2b+7)^2 - (2a)^2 = 49 =>(2b+2a+7)(2b - 2a+7)=49
FriIzi
08.10.2022 15:48
My solution: a²=b²+7b We can get it to quadratic formula We can multiply RHS and LHS to 4 And 4a²=4b²+28b We can add 49 to RHS and we must do -49 4a²=4b²+28b+49-49 4a²=(2b+7)²-49 So 4a²-(2b+7)²=-49 We can LHS and RHS mutiply by -1 and (2b+7)²-4a²=49 And a²-b²=(a-b)(a+b) By forumla (2b+7-2a)(2b+7+2a)=49 1. 2b+7-2a={1,7} 2. 2b+7+2a={7,49} If 2b-2a+7=7 2b-2a=0 B=a And 4a+7=7 (B=a=0) first solution The second one is 2b-2a=6 B-a=3 B=3+a 4a+13=49 4a=36 and (a=9 b=12) And we obtain 2 solution
mathmax12
06.09.2023 14:46
Using, the quadratic, formula, we need, $b=\frac{-7 \pm \sqrt{49+4a^2}}{2}$, so $49+4a^2=b^2$, for a odd integer, $b.$ Rest, is just casework, bash(have to go to school).