Finding all integers with a divisibility condition

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Finding all integers with a divisibility condition

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Source: Germany 2020, Problem 4

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#1 h Jun 22, 2020, 4:09 PM • 2 Y

Y by Math_olympics, itslumi

Determine all positive integers $n$ for which there exists a positive integer $d$ with the property that $n$ is divisible by $d$ and $n^2+d^2$ is divisible by $d^2n+1$ .

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#2 h Jun 22, 2020, 4:32 PM • 2 Y

Y by Math_olympics, zhihanpeng

The answers are all positive integers $k^4$ which is true by taking $d = k$ .
We'll now prove that there are no other integers that have this property.
Let $n = da$ for an integer $a$ . We then have
$d^3 a + 1 \mid a^2 + 1$ From here, by size reasons we get that $a \ge d^3$ .
But notice that
$d^3 a + 1 \mid d^6(a^2 + 1) - (d^6 a^2 - 1) = d^6 + 1$ Therefore, $a \le d^3$ .
This forces $a = d^3$ .

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ayan_mathematics_king

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ayan_mathematics_king

#3 h Dec 19, 2020, 2:31 PM • 2 Y

Y by Math_olympics, amar_04

How about this?

We have $d^3a+1|a^2+1 \implies d^3a+1|a^2+1-d^3a-1$
Therefore, $d^3a+1 | a^2-d^3a$
$\implies d^3a+1 |a(a-d^3) \implies d^3a+1|a-d^3$ .

Case-1: $a< d^3$
Then, $d^3a+1 \le d^3-a$
$\implies a(d^3+1)\le d^3-1$ $a\le \frac{d^3-1}{d^3+1}<1$
Which is absurd.

Case-2: $a >d^3$
Then, $d^3+1\le a-d^3$
$\implies a(d^3-1)\le -d^3-1$ Therefore, $a\le \frac{-(d^3+1)}{d^3-1}$ .
Which is not possible since $a$ is positive.

So, clearly $a=d$ is the possible case. Fortunately, $x=d^4$ works, so we are done.

This post has been edited 3 times. Last edited by ayan_mathematics_king, Dec 19, 2020, 3:33 PM

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#4 h Dec 19, 2020, 2:37 PM • 1 Y

Y by Math_olympics

I'm a little confused on how you got $d^3a+1 \mid a^2+1$

This post has been edited 3 times. Last edited by eduD_looC, Dec 19, 2020, 2:38 PM
Reason: why should be how

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#5 h Dec 19, 2020, 2:47 PM • 2 Y

Y by Math_olympics, Mango247

Well, by assumption we have $d^3a+1 \mid d^2+n^2=d^2(a^2+1)$ . But $d^3a+1$ is clearly coprime to $d$ and hence to $d^2$ so that this implies $d^3a+1 \mid a^2+1$ .

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#6 h Dec 19, 2020, 2:51 PM • 3 Y

Y by Math_olympics, Mango247, Mango247

Oh, thank you @Tintarn.

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#7 h Dec 19, 2020, 3:03 PM • 1 Y

Y by Math_olympics

I also don't get how $d^3a+1 \le d^3-a$ in Case 1.

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#8 h Dec 19, 2020, 3:11 PM • 1 Y

Y by Math_olympics

eduD_looC wrote:

I also don't get how $d^3a+1 \le d^3-a$ in Case 1.

Yes, this is the actual critical point. This woks if $a<d^3$ since then $d^3-a$ is a positive multiple of $d^3a+1$ and hence at least as large as $d^3a+1$ . But of course this does not work if $d^3=a$ .
So in fact the solution in #3 would be correct by noting that this is the only missing case.

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ayan_mathematics_king

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ayan_mathematics_king

#9 h Dec 19, 2020, 3:32 PM • 2 Y

Y by Math_olympics, amar_04

Tintarn wrote:

eduD_looC wrote:

I also don't get how $d^3a+1 \le d^3-a$ in Case 1.

Yes, this is the actual critical point. This woks if $a<d^3$ since then $d^3-a$ is a positive multiple of $d^3a+1$ and hence at least as large as $d^3a+1$ . But of course this does not work if $d^3=a$ .
So in fact the solution in #3 would be correct by noting that this is the only missing case.

Ok thanks...I figured out my mistake.

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#10 h Jun 26, 2021, 1:00 AM

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One straightforward solution: consider the problem in the modulus $d^2n+1$ , which means that we can change $n$ to $- \dfrac{1}{d^2}$ and then the problem is rewritten.(Write $n$ as $dn_1$ and as $n \equiv -\dfrac{1}{d^2} (\mod nd^2+1)$ , we have $n_1d^3+1 | d^3-n_1$ . Compare the left and right hand side and we'll get $n_1=d^3 \Rightarrow n=dn_1=d^4$ .) :-D

:-D

This post has been edited 1 time. Last edited by MathxSudio, Jun 26, 2021, 12:13 PM
Reason: to add some details

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#11 h Jun 26, 2021, 6:45 AM

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Fair enough, but how is "rewriting the problem" a "straightforward solution"?
I fail to see how the problem is trivial after that rewriting (of course it's not hard either, but then again the whole problem was not very hard to begin with...).

This post has been edited 1 time. Last edited by Tintarn, Jun 26, 2021, 6:46 AM

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L13832

#12 h Jul 13, 2024, 7:57 AM

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Let $n=dk$ where $n,d,k \in \mathbb{N}$ . Note that $\begin{align*}d^2n+1 \mid n^2+d^2 &\implies d^3k+1 \mid d^2(k^2+1) \implies d^3k+1 \mid k^2+1 \\& \implies d^3k+1 \leq k^2+1 \implies k \geq d^3 \qquad (*)\end{align*}$ Now because $\begin{align*} d^3k+1 \mid k^2+1& \implies d^3k+1 \mid k^2+1-d^2k-1=k^2-kd^3 \\& \implies k-d^3\geq d^3k+1\end{align*}$ But $d^3k+1 \geq k-d^3$ so $k-d^3=0 \implies k=d^3.$
Therefore, $\boxed{n=d^4}$ where $d \in \mathbb{N}$ .

This post has been edited 2 times. Last edited by L13832, Sep 15, 2024, 8:45 AM

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#13 h Mar 20, 2025, 5:44 AM

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Tintarn wrote:

Determine all positive integers $n$ for which there exists a positive integer $d$ with the property that $n$ is divisible by $d$ and $n^2+d^2$ is divisible by $d^2n+1$ .

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#14 h Apr 24, 2025, 4:54 PM

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Assumption: n=dk

Claim 1: $n=d^{4}$

Proof 1 : $d^{3}k +1 | d^{2}(k^{2} + 1)$ => $d^{3}k +1 | k^{2} + 1$ As gcd(d^{3}k+1,d^{2})
=> $d^{3}k +1 \leq k^{2} + 1$ => $d^{3} \leq k$
Also $d^{3}k +1 | d^{2}(k^{2} + 1)$ => $d^{3}k +1 | d^{3}(k^{2} + 1)$ => $d^{3}k +1 | d^{3}(k^{2} + 1) - k(d^{3}k +1)$ => $d^{3}k +1 | d^{3} -k$

Claim 1.1 : $d^{3} = k$

Proof 1.1 : FTSOC $d^{3}$ < $k$
$d^{3}k +1 \leq k-d^{3}$ => $d^{3}k$ < $k-d^{3}$ => $d^{3}(k+1)$ < $k$ => $d^{3}$ < $\frac{k}{k+1}$ < 1 nonsense

Thus $d^{3} = k$
Thus $d^{4} = n$

Remark: Easy problem

I rate it d3

Solved: 16/4/25

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#15 h Apr 24, 2025, 10:06 PM

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Almost the same as Bulgaria Winter Mathematical Competition 2013, Junior Level.

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#16 h Apr 30, 2025, 3:19 AM

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Let $n=ds$ , then $d^3s|s^2+1$ , so we have $s\geq d^3$
if $d^3<s$ then we have $d^3s+1|s^2+1$ , then $d^3s+1|s^2-d^3s$
So $s-d^3\geq d^3s+1$ .
By the two ineq we have $s=d^3$ and so $n=d^4$ .

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