Find the set of all ordered pairs $(s,t)$ of positive integers such that \[t^{2}+1=s(s+1).\]
Problem
Source: Turkish Mathematical Olympiad 2nd Round 1994
Tags: quadratics, calculus, integration, number theory unsolved, number theory
Rust
27.09.2006 15:09
Let $x=2s+t$ then $x^{2}-5t^{2}=4$, therefore $x=q^{n}+q^{-n}, \ q=9+4\sqrt 5 ,q^{-1}=9-4\sqrt 5 ,t=\frac{1}{\sqrt 5 }(q^{n}-q^{-n})$ and $s=\frac{x-t}{2}$ (x and t are even.
xeroxia
14.07.2012 04:19
I have changed the problem statement from $ t^{2}+1=s(s+t) $ to $ t^{2}+1=s(s+1) $. http://www.imocompendium.com/othercomp/Tur/TurMO94.pdf says it is $t$. A Turkish monthly math journal says it is $1$.
mathbuzz
14.07.2012 11:00
s(s+1)=$t^2+1$ . so , $s^2+s-(t^2+1)$=0. now we treat it as a quadratic in s. for having integral solution , we must have that D=$1-4.1.-(t^2+1)$=$(1+4[t^2+1])$=$4t^2+5$ is a perfect square. so .let $4t^2+5=y^2$ , which is trivial to solve using standard factorization.
pablock
20.12.2017 20:54
$(s-1)^2 < t^2 = s(s+1)-1 < (s+1)^2 \implies t^2=s^2=s(s+1)-1 \implies (s,t)=(1,1)$.
GRCMIRACLES
06.01.2018 10:08
multiply the equation by 4,we get 4t^2+4=4s^2+4s =>{2s+1}^2-{2t}^2=5 =>{2s+2t+1}{2s-2t+1}=5.1 =>2s+2t+1=5,2s-2t+1=1 =>s=t=1 is the only solution in positive integers
Eyed
14.01.2020 06:15
If $t,s\geq 2$, then we can see the following:
If $s\geq t$, then $s(s+1)\geq t(t+1) = t^2 + t>t^2+1$.
Next, if $s<t$, then $s(s+1)<(t-1)(t) = t^2-t<t^2+1$.
Therefore $t,s$ cannot be greater or equal to $2$.
If $s$ was $1$, then we have $t^{2} + 1 = 2$, so $t=1$. Similarly, if $t=1$, then we also get that $s = 1$. Therefore, our only solution is $(s,t) = (1,1)$.
reeh_haan
01.07.2022 14:31
$t^2+1=s\left(s+1\right)$
$t^2+1=s^2+s$
$t^2+1+\frac 14=s^2+s+\frac 14$
$t^2+\frac 54=\left(s+\frac 12\right)^2$
$\frac 54=\left(s+\frac 12\right)^2-t^2$
$\frac 54=\left(s+t+\frac 12\right)\left(s-t+\frac 12\right)$
$5=\left(2s+2t+1\right)\left(2s-2t+1\right)$
$\Longrightarrow 2s+2t+1=5, 2s-2t+1=1\implies (s, t)=\boxed{(1, 1)}$
$\Longrightarrow 2s+2t+1=-5, 2s-2t+1=-1\implies (s, t)=(-2, -1)\notin\mathbb N$
$\Longrightarrow 2s+2t+1=1, 2s-2t+1=5\implies (s, t)=(1, -1)\notin\mathbb N$
$\Longrightarrow 2s+2t+1=-1, 2s-2t+1=-5\implies (s, t)=(-2, 1)\notin\mathbb N$
turkOklid
12.12.2024 10:20
$t^2+1=s^2+s \Rightarrow t^2-s^2=(t-s)(t+s)=s-1$ if both sides of the equation are positive, $t+s \leq s-1 \Rightarrow t\leq -1$ But t is positive, contradiction. Then $s \geq t$, $(s-t)(s+t)=1-s$ On this equation left hand side is non-negative then right hand side should be non-negative too. Since $s$ is positive both side should be zero. Then $(s,t)=(1,1)$ is only solution.