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Source: Bulgaria National Olympiad 2025, Day 1, Problem 1

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#1 h Apr 8, 2025, 1:47 PM • 2 Y

Y by PikaPika999, cubres

Determine all infinite sequences $a_1, a_2, \ldots$ of real numbers such that
$a_{m^2 + m + n} = a_{m}^2 + a_m + a_n$ for all positive integers $m$ and $n$ .

This post has been edited 2 times. Last edited by Assassino9931, Apr 9, 2025, 10:32 PM

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#2 h Apr 8, 2025, 2:49 PM • 2 Y

Y by PikaPika999, cubres

Retracted since problem was changed/corrected.

This post has been edited 1 time. Last edited by alexheinis, Apr 9, 2025, 8:47 AM

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#3 h Apr 8, 2025, 4:53 PM • 2 Y

Y by PikaPika999, cubres

@above Apologies, I fixed the problem statement - it should be $a_m^2$ , not $a_{m^2}$ .

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Primeniyazidayi

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Primeniyazidayi

#4 h Apr 8, 2025, 5:27 PM • 2 Y

Y by PikaPika999, cubres

Assassino9931 wrote:

@above Apologies, I fixed the problem statement - it should be $a_m^2$ , not $a_{m^2}$ .

One can put m=1 and note that $a_k=\frac{k-1}{2}a_1^2 + \frac{k+1}{2}a_1$ for odd k.One can now put n=1 to see that $\frac{m^2+m}{2}a_1^2 + \frac{m^2+m+2}{2}a_1=a_m^2+a_m+a_1$ or $a_m^2+a_m=\frac{m^2+m}{2}(a_1^2+a_1)$ for all m.Let $b_n=\frac{a_m^2+a_m}{m^2+m}$ .Then it is trivial that $b_n$ is constant,which means that $a_m^2+a_m-C(m^2+m)=0$ for some fixed constant C.I think that rest is not so hard.

This post has been edited 1 time. Last edited by Primeniyazidayi, Apr 8, 2025, 5:28 PM

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#5 h Apr 8, 2025, 7:01 PM • 2 Y

Y by PikaPika999, cubres

$a_1=a,a_2=b$
$a_{n+2}=a_n+a^2+a$ so $a_{n+2k}=a_n+k(a^2+a)$

$a_{n+m^2+m}=a_n+ \frac{(m^2+m)(a^2+a)}{2} \to a_m^2+a_m= \frac{(m^2+m)(a^2+a)}{2}$

$a_3=a^2+2a$
$a_3^2+a_3= (a^2+2a)^2+(a^2+2a)=6(a^2+a) \to a^4+4a^3-a^2-4a=0 \to a=-4,-1,0,1$

$a_5=2a^2+3a$
$a_5^2+a_5=(2a^2+3a)^2+(2a^2+3a)=15(a^2+a) \to a= -1,0,1$

$a_{n+6}=a_n+b^2+b$ so $3(a^2+a)=b^2+b$

$a_4=a^2+a+b$
$a_{n+20}=a_4^2+a_4+a_n$ so $a_4^2+a_4=(a^2+a+b)^2+(a^2+a+b)=10(a^2+a)$

$a=-1,0 \to b=0,-1$ and $a_{1+2k}=a, a_{2+2k}=b$
$a=1 \to b=2$ and $a_{1+2k}=1+2k, a_{2+2k}=2+2k$

So answer is $(-1,0,-1,0,...), (0,0,0,0,...),(-1,-1,-1,-1,...),(0,-1,0,-1,...), (1,2,3,4,...)$

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#6 h Apr 9, 2025, 12:21 AM • 3 Y

Y by PikaPika999, AlexCenteno2007, cubres

Let $a_1=\alpha$ then notice $a_{n+2}=a_n+\alpha^2+\alpha$ so we get $a_{n+2k}=a_n+k(\alpha^2+\alpha)$ and so this gives that $a_{m^2+m+n}=a_n+\frac{m^2+m}{2} \cdot (\alpha^2+\alpha)$ which now gives $\frac{a_m^2+a_m}{m^2+m}=\frac{\alpha^2+\alpha}{2}$ .
Now from here notice we get $2a_m^2+2a_m+2(\alpha^2+\alpha)(2a_m+\alpha^2+\alpha+1)=(\alpha^2+\alpha)(m^2+m)+(\alpha^2+\alpha)(4m+6)$ which gives that either $\alpha=0,-1$ or $4a_m+2(\alpha^2+\alpha+1)=4m+6$ which shows that $a_m=m+\theta$ for some real $\theta$ and plugging this back would give that $\frac{2m \theta+\theta^2+\theta}{m^2+m}=\frac{\alpha^2+\alpha-2}{2}$ and clearly this doesn't hold true for all positive integers $m$ unless $\theta=0$ in which case $a_m=m$ for all positive integers $m$ which indeed works.
Now in bith other cases $\alpha^2+\alpha=0$ and thus $a_{m+2}=a_m$ and $a_m(a_m+1)=0$ always hold which means that $a_m \in \{-1,0 \}$ for all positive integers $m$ , it is easy to see from both cases now that only spanned solutions can be $a_m=0$ always or $a_m=-1$ always or $a_{2k}=-1$ and $a_{2k-1}=0$ or $a_{2k}=0$ and $a_{2k-1}=-1$ , these four solutions also work perfectly thus we are done :cool:

:cool:

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This post has been edited 1 time. Last edited by MathLuis, Apr 9, 2025, 12:22 AM

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#7 h Apr 9, 2025, 8:23 PM

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Let us denote $a_1 = x$ . We have $a_{n+2} = a_n + x^2 + x$ , in particular $a_3 = x^2 + 2x$ and $a_{13} = a_1 + 6(x^2 + x) = 6x^2 + 7x$ . On the other hand, $a_{13} = a_3^2 + a_3 + a_1 = (x^2 + 2x)^2 + (x^2 + 2x) + x = x^4 + 4x^3 + 5x^2 + 3x$ . By equating the two expressions, we get $x^4 + 4x^3 - x^2 - 4x = 0$ , which factors as $x(x-1)(x+1)(x+4) = 0$ , so $a_1 = 0, 1, -1$ or $-4$ .

If $a_1 = -4$ , then $a_{n+2} = a_n + 12$ for every $n$ , and $a_5 = 20$ , $a_{35} = a_1 + 17 \cdot 12 = 200$ , but also $a_{35} = a_5^2 + 2a_5 = 440$ , which is a contradiction.

Now let $a_2 = y$ . From above we have $a_7 = a_1 + 3(x^2 + x) = 3x^2 + 4x$ , and on the other hand $a_7 = a_2^2 + a_2 + a_1 = y^2 + y + x$ , so $y^2 + y = 3x^2 + 3x$ . We now consider the following cases:

- If $x = 0$ or $x = -1$ , then $y = 0$ or $y = -1$ , and $a_{n+2} = a_n$ for all $n$ . All four possibilities work, since $t^2 + t = 0$ for $t = 0, -1$ , and $m^2 + m + n$ and $n$ have the same parity.

- If $x = 1$ , then $y^2 + y = 6$ , so $y = 2$ or $y = -3$ ; also $a_{n+2} = a_n + 2$ . In the first case, by induction we get $a_n = n$ for every $n$ (which clearly satisfies the given condition). In the second case, $a_4 = -1$ , $a_{21} = a_4^2 + a_4 + a_1 = 1$ , but also $a_{21} = a_1 + 2 \cdot 10 = 21$ , a contradiction.

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