Turkish MO 1994 P5

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Turkish MO 1994 P5

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Source: Turkish Mathematical Olympiad 2nd Round 1994

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1133 posts

#1 h Sep 27, 2006, 10:06 AM • 2 Y

Y by Adventure10, Mango247

Find the set of all ordered pairs $(s,t)$ of positive integers such that $t^{2}+1=s(s+1).$

This post has been edited 2 times. Last edited by xeroxia, Apr 7, 2013, 8:37 AM

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#2 h Sep 27, 2006, 12:09 PM • 2 Y

Y by Adventure10, Mango247

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.

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#3 h Jul 14, 2012, 1:19 AM • 3 Y

Y by Pluto1708, Adventure10, Mango247

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 $1$ .

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#4 h Jul 14, 2012, 8:00 AM • 1 Y

Y by Adventure10

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.

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168 posts

#5 h Dec 20, 2017, 5:54 PM • 2 Y

Y by pavel kozlov, Adventure10

$(s-1)^2 < t^2 = s(s+1)-1 < (s+1)^2 \implies t^2=s^2=s(s+1)-1 \implies (s,t)=(1,1)$ .

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#6 h Jan 6, 2018, 7:08 AM • 2 Y

Y by Adventure10, Mango247

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

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#7 h Jan 14, 2020, 3:15 AM • 2 Y

Y by kevinmathz, Adventure10

Sol

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#8 h Jul 1, 2022, 11:31 AM

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#9 h Dec 12, 2024, 7:20 AM

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$t^2+1=s^2+s \Rightarrow t^2-s^2=(t-s)(t+s)=s-1$
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.

This post has been edited 1 time. Last edited by turkOklid, Dec 12, 2024, 7:21 AM

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Primeniyazidayi

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Primeniyazidayi

#10 h Apr 3, 2025, 7:22 AM

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Extremely easy:
For some $a$ we should have $4t^2 + 5 =a^2$ and 1 second casework finishes.

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