An equation from the past with different coefficients

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An equation from the past with different coefficients

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Source: Balkan MO Shortlist 2024 N2

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1384 posts

#1 h Apr 27, 2025, 1:00 PM • 1 Y

Y by PikaPika999

Let $n$ be an integer. Prove that $n^4 - 12n^2 + 144$ is not a perfect cube of an integer.

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#2 h Apr 27, 2025, 1:02 PM • 1 Y

Y by PikaPika999

Similar to HMO 2025 p4.

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1384 posts

#3 h Apr 27, 2025, 1:05 PM • 1 Y

Y by PikaPika999

@above Indeed, the technique from there is useful (and popular), but the very essential similarity is with this JBMO Shortlist 2020 problem.

This post has been edited 1 time. Last edited by Assassino9931, Apr 27, 2025, 1:05 PM

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228 posts

#4 h Apr 27, 2025, 1:06 PM

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Also AZE BMO TST P3

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MR.1

#5 h Apr 27, 2025, 1:11 PM

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GreekIdiot wrote:

Similar to HMO 2025 p4.

what is hmo?

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228 posts

#6 h Apr 27, 2025, 1:14 PM

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MR.1 wrote:

GreekIdiot wrote:

Similar to HMO 2025 p4.

what is hmo?

Hellenic Math Olympiad

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114 posts

#7 h Apr 27, 2025, 1:15 PM

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Similar to Greece Math olympiad 2025

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Tamam

#8 h Apr 27, 2025, 1:23 PM • 1 Y

Y by Lahmacuncu

$n^4-12n^2+144=(n^2-6n+12)(n^2+6n+12)$ from Euclidian Algorithm we can deduce that $gcd(n^2-6n+12,n^2+6n+12)=(n,12)$
If $n$ is even let $n=2k$ we will get that $16(k^4-3k^2+9)=m^3$ from here we know $v_2$ of $LHS$ is 4 which is not divisible by 3
If $n$ is divisible by 3 let $n=3k$ we get $9(9k^4-12k^2+1)=m^3$ since m is divisivible by 3 $LHS$ should be divisible by 27 which yields a contradiction.
So both the terms in the product are perfect cubes.
Now let $a=n-3$ which is even. Let $(n-3)^2+3=n^2-6n+12=b^3$ . So $a^2+4=b^3+1=(b+1)(b^2-b+1)$ Since $b\equiv 3 (mod 4)$ there exists a prime congurent to 3 modulo 4 such that $p|b^2-b+1|a^2+2^2$ but from this we get $p=2$ which is a contradiction. So no solution.

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114 posts

#9 h Apr 27, 2025, 1:27 PM

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My solution in the tst

Attachments:

BMO 2024 SL N2.pdf (76kb)

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932 posts

#10 h Apr 27, 2025, 2:01 PM • 3 Y

Y by MuradSafarli, farhad.fritl, ehuseyinyigit

This problem was proposed by me (Dorlir Ahmeti, Albania).

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5611 posts

#11 h Apr 27, 2025, 3:55 PM

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Suppose the expression was a cube for some $n$ . Factor as $(n^2 - 6n + 12)(n^2 + 6n + 12)$ .

Claim: $n^2 - 6n + 12$ and $n^2 + 6n + 12$ are relatively prime.
Proof: Suppose otherwise. Firstly, if $p$ divides both $n^2 - 6n + 12$ and $n^2 + 6n + 12$ , then $p$ divides $12n$ , and since $p$ divides $12(n^2 - 6n +12)$ , $p$ divides $144$ , so $p$ is $2$ or $3$ . But if $p = 3$ , then $3 \mid n$ , but this means $n^2 - 6n + 12$ and $n^2 + 6n + 12$ are both $12 \equiv 3 \pmod 9$ , so their product has a $\nu_3$ of $2$ , so it is not a perfect cube. Thus, $\gcd(n^2 - 6n + 12, n^2 + 6n + 12)$ is a power of $2$ . Since the gcd is not $1$ , it is even, so $2$ divides $n$ . Since $n$ and $n - 6$ are even and different mod $4$ , $8 \mid n(n-6) = n^2 - 6n$ , so since $8 \mid 12n$ , $8$ also divides $n^2 - 6n + (12n) = n^2 + 6n$ . Thus, $n^2 - 6n + 12$ and $n^2 + 6n + 12$ are $12 \equiv 4 \pmod 8$ , so their product has a $\nu_2$ of $4$ , so it is not a perfect cube. $\square$

Thus, $n^2 - 6n + 12$ and $n^2 + 6n + 12$ are perfect cubes. Taking modulo $7$ gives that $n^2 - n + 5$ and $n^2 + n + 5$ are both in $\{-1,0,1\}$ modulo $7$ . Multiplying by $4$ and taking mod $7$ gives $(2n-1)^2 - 2$ and $(2n+1)^2 - 2$ are in $\{-4,0,4\}\pmod 7$ .

Hence $(2n-1)^2$ and $(2n+1)^2$ are in $\{-2,2,6\}\pmod 7$ , and since $-2$ and $6$ are not QRs modulo $7$ , they are both $2$ modulo $7$ .

Thus, $(2n+1)^2 - (2n-1)^2 = 8n \equiv 0 \pmod 7$ , so $7 \mid n$ , but then $(2n-1)^2 \equiv 1 \pmod 7$ , absurd. Therefore, no such integer $n$ exists.

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636 posts

eg4334

#12 h Apr 27, 2025, 5:07 PM

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First, factor it as $(n^2-6n+12)(n^2+6n+12)$ . The only prime factors they can share are $2$ or $3$ , because they must divide $12n$ . If they are even, then $n=2a$ so we need $16(a^4-3a^2+9)$ to be a cube which is impossible because the parenthesses is odd. Similarly, if they are divisible by $3$ then $n=3a$ but $9(9a^4-12a^2+16)$ cannot be a cube by $v_3$ analysis. Therefore for the sake of our problem they are relatively pirme. Then let $a^3=n^2-6n+12, (a+3x)^3 = n^2+6n+12$ , because their difference must be a multiple of $3$ . Then $12n=9a^2x+27ax^2+27x^3 \implies n = \frac{3a^2x+9ax^2+9x^3}{4}$ . Put this into the latter equation to get $16(a+3x)^3 = (3a^2x+9ax^2+9x^3)^2 + 24(3a^2x+9ax^2+9x^3)+192$ . Its easy to see that when $x \geq 2$ , this is impossible by size constraints because the right side is so big. When $x=1$ , we have $12n = 9a^2+27a+27$ which is impossible by $\pmod{4}$ . Therefore there is no such $n$ ,.

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ektorasmiliotis

111 posts

ektorasmiliotis

#13 h Apr 27, 2025, 7:48 PM

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in HMO: x^2 +108= is not a perfect cube

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#14 h Apr 27, 2025, 10:13 PM

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Set $n^4-12n^2+144=k^3$ . First, I prove $k\equiv 0,3\pmod{4}$ are impossible. Suppose $4\mid k$ and $k=4u$ . Notice that $v_2(LHS)=4$ for $v_2(n)\ge 2$ , so $v_2(n)=1$ necessarily. Setting $n=2t$ for $t$ odd, we find $16t^4 - 48t^2 + 144 = 64u^3\Rightarrow t^4-3t^2+9 = 4u^3$ which is impossible by parity. Likewise, if $k\equiv 3\pmod{4}$ , then $n^4-12n^2+144\equiv n^4\equiv 3\pmod{4}$ , a contradiction.

In the remainder, we will rule out $k\in\{1,2\}\pmod{4}$ . Suppose first $k\equiv 1\pmod{4}$ . Note that
$(n^2-6)^2 + 10^2 = k^3 - 8 =(k-2)(k^2+2k+4).$ Since $k-2\equiv -1\pmod{4}$ , there is a prime $q\equiv 3\pmod{4}$ such that $q\mid \alpha^2+\beta^2$ for $\alpha=n^2-6$ and $\beta = 10$ . It is well known that $q\mid\alpha,\beta$ in this case, but $q\mid 10$ is clearly impossible.

Lastly, suppose $k\equiv 2\pmod{4}$ . We have
$(n^2-6)^2 + 9^2 = k^3 - 27 = (k-3)(k^2+3k+9).$ Once again, $k-3\equiv -1\pmod{4}$ , so there is a prime $p\equiv 3\pmod{4}$ dividing both $n^2-6$ and $9$ . At the same time, $3\mid n$ is impossible: otherwise $v_3(n^4)\ge 4$ , $v_3(12n^2)\ge 3$ whereas $v_3(144)=2$ gives $v_3(LHS)=2$ , not possible for a perfect cube.

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