Cubes and squares

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y-is-the-best-_

16 posts

y-is-the-best-_

#1 h Sep 22, 2020, 11:32 PM • 5 Y

Y by itslumi, megarnie, veirab, deplasmanyollari, cubres

Find all triples $(a, b, c)$ of positive integers such that $a^3 + b^3 + c^3 = (abc)^2$ .

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pad

1671 posts

pad

#2 h Sep 22, 2020, 11:33 PM • 5 Y

Y by mathingbingo, guptaamitu1, wasikgcrushedbi, bhan2025, cubres

WLOG $a\le b\le c$ . Since $c^2$ divides the RHS, it also divides $a^3+b^3+c^3$ , hence $c^2\mid a^3+b^3$ ; in particular, $c^2\le a^3+b^3$ . Also, $3c^3 \ge (abc)^2$ , so $3c\ge a^2b^2$ , which means $c^2 \ge \tfrac{a^4b^4}{9}$ . Now,
$\frac{a^4b^4}{9} \le a^3+b^3 \implies a^4b \le 9\left(\frac{a^3}{b^3}+1\right)\le 9\cdot 2= 18.$ This clearly implies $a=1$ . The equation is now $1+b^3+c^3=b^2c^2$ . So $2c^3+1 \ge b^2c^2$ . If $2c^3+1=b^2c^2$ , then $c^2(b^2-2c)=1$ , so $c=1$ , contradiction. Therefore, $2c^3 \ge b^2c^2$ , so $c\ge \tfrac12b^2$ . Also, $c^3 \le b^2c^2$ , so $c\le b^2$ . In summary, $c\in \left[ \tfrac12 b^2, b^2\right]$ .

Suppose $b\ge 5$ . Then $c^2\ge \tfrac14 b^4 \ge b^3+31$ . This means $(b^2-c)c^2 \ge b^3+31$ (assuming $c\not = b^2$ , but this case is easy to deal with). Hence $b^2c^2 \ge b^3+c^3+31$ , contradiction.

Therefore, $b\le 4$ , and bashing out these cases gives the only solution as $(b,c)=(2,3)$ . In conclusion, the answers are $(a,b,c)=(1,2,3)$ and permutations.

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v_Enhance

6877 posts

v_Enhance

#3 h Sep 22, 2020, 11:36 PM • 16 Y

Y by Mathematicsislovely, Pluto1708, Math_olympics, v4913, Kontrol, math31415926535, Slavici_Adrian, HamstPan38825, Spinglasses, rayfish, MathxSudio, veirab, sabkx, bhan2025, cubres, Kingsbane2139

The answer is $(1,2,3)$ and permutations which works. Assuming $a \le b \le c$ , write the given equation as $a^3 + b^3 = c^2(a^2b^2-c).$ Thus actually assume $a \le b \le c < a^2b^2$ .
Write $\begin{align*} 2b^3 &\ge a^3+b^3 = c^2(a^2b^2-c) \\ &\ge \min \left\{ b^2(a^2b^2-b), (a^2b^2-1)^2 \right\} \\ &\ge \min \left\{ b^2(b^2-b), (b^2-1)^2 \right\}. \end{align*}$ where the second equality follows from the fact $x \mapsto x^2(a^2b^2-x)$ has no local minima for $0 < x < a^2b^2$ , hence is minimized at endpoints.
This implies $b \le 3$ , so we just need to manually examine $1 \le a \le b \le 3$ for solutions in $c$ . Doing so gives the solution.

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anyone__42

92 posts

anyone__42

#4 h Sep 22, 2020, 11:37 PM • 1 Y

Y by cubres

https://artofproblemsolving.com/community/c6h1749156p11418402

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dangerousliri

931 posts

dangerousliri

#5 h Sep 22, 2020, 11:44 PM • 1 Y

Y by cubres

https://artofproblemsolving.com/community/c6h359192p1962933

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trumpeter

3332 posts

trumpeter

#6 h Sep 23, 2020, 12:36 AM • 5 Y

Y by N0_NAME, ayan_mathematics_king, IAmTheHazard, veirab, cubres

The answers are $(1,2,3)$ and permutations. WLOG let $a\geq b\geq c$ .

Suppose that $b\geq3$ . Define the function $f(t)=t^3-b^2c^2t+b^3+c^3$ so that $a$ is a root of $f$ . Observe that:
$\begin{align*} f(-\infty) &= -\infty < 0 \\ f(0) &= b^3+c^3 > 0 \\ f(b) &= -b^3(bc^2-3)-(b^3-c^3) < 0 \\ f(b^2c^2-1) &= -2b^3(\frac{32}{81}b^2c^4-1)-(\frac{1}{9}b^2c^2-1)(\frac{17}{9}b^2c^2-1)-(b^3-c^3) < 0 \\ f(b^2c^2) &= b^3+c^3 > 0 \end{align*}$ By the intermediate value theorem, $f$ has a root in each of the intervals $(-\infty,0),(0,b),(b^2c^2-1,b^2c^2)$ . Since $f$ is a cubic polynomial, it no other roots. But then $f$ has no integer root at least $b$ , which is a contradiction.

Thus $b\leq2$ . Check that the only solution is $(a,b,c)=(3,2,1)$ as desired.

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dchenmathcounts

2443 posts

dchenmathcounts

#7 h Sep 23, 2020, 4:46 AM • 2 Y

Y by Mango247, cubres

Without loss of generality say $a\leq b\leq c.$ This is an inequality. Check that $(1,1,c)$ does not work in general to avoid sadness.

Note $c^2\mid a^3+b^3,$ so $c^2\leq a^3+b^3.$ Now note that we want
$(abc)^2\leq a^3+b^3+c^3\leq (a^3+b^3)(1+c)$ $(ab)^2c< 2a^3+2b^3, \text{ since } 1+c>2$ $a^2b^3<4b^3$ $a^2<4,$ implying that we want $a<2,$ or in particular, $a=1.$

Now we rinse and repeat. We want
$(bc)^2\leq 1+b^3+c^3\leq (1+b^3)(1+c).$ We claim that equality cannot hold unless $b=2$ ; note that otherwise $b\mid 1+c$ or $c\geq 2b-1,$ epicly failing for size reasons since this implies $c\geq \frac{3}{2}b,$ and if $b> 2$ and $k=\frac{c}{b}\geq \frac{3}{2},$
$b^4k^2\geq 1+b^3+bk+b^4k,\text{ since}$ $bk(b^3k-b^3-1)\leq b^3(k-1)\geq bk\left(\frac{b^3}{2}-1\right)\geq \frac{25}{54}b^4k \text{ and }$ $b^3+1\leq \frac{28}{27}b^3, \text{ and}$ $25bk\geq 56,$ contradiction.

Thus we want
$(bc)^2\leq b^3c+b^3+c$ $c(b^2c-1)\leq b^3(c+1)$ and note that $b^2c-1\geq \frac{35}{27}b^3$ as we assume $b\geq 3$ and obviously $b\neq c,$ and that $c+1\leq \frac{5}{4}{c}.$ So we have
$\frac{35}{27}b^3c\leq \frac{5}{4}b^3c,$ which is ridiculous, so we reach a contradiction.

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square_root_of_3

78 posts

square_root_of_3

#8 h Sep 23, 2020, 9:10 AM • 4 Y

Y by chengalexzhongzhi, Mano, Carius_MrPenguin, cubres

Here's a cool solution I found. Without loss of generality say $a \geqslant b \geqslant c$ . We know that $a$ is a root of the polynomial $x^3-(bc)^2x^2+b^3+c^3$ . Let $u$ and $v$ be the other two roots. By Vieta's formulas, we have $a+(u+v)=b^2c^2,$ $a(u+v)=-uv,$ $(-uv)a=b^3+c^3.$ From the first equation, we see that $u+v$ is an integer, and from the second and third equation we see that $u+v>0$ . Let $w=u+v$ . Then, eliminating $uv$ , we obtain the following equations: $a+w=b^2c^2,$ $aw=\frac{b^3+c^3}{a}.$ Using the inequality $aw+1\geqslant a+w$ , we have $b^2c^2\leqslant \frac{b^3+c^3}{a}+1 \leqslant 2b^2+1 \implies c=1,$ where we used the ordering $a\geqslant b \geqslant c$ for the last inequality.

The equations are now $a+w=b^2,$ $aw=\frac{b^3+1}{a}\leqslant b^2+1.$ If $w\geqslant 2$ and $a>2$ , we have $(a-1)(w-1)\geqslant 2$ , which implies $aw>a+w+2 \implies b^2+1>b^2+2$ , a contradiction.

If $w=1$ , then $a=b^2-1$ and $a^2=b^3+1$ , which implies $b^2(b^2-2)=b^3 \implies b=2, a=3$ , which is a solution.

If $a\leqslant 2$ , then the only triples we need to check are $(1,1,1)$ , $(2,1,1)$ and $(2,2,1)$ . Since none of them works, the only solution is $(3,2,1)$ .

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MarkBcc168

1595 posts

MarkBcc168

#9 h Sep 23, 2020, 9:35 AM • 9 Y

Y by Math-wiz, Jupiter_is_BIG, Kanep, Mathmick51, guywholovesmathandphysics, rightways, veirab, GeoKing, cubres

All answers are the six permutations of $(3,2,1)$ .

WLOG, let $a\geq b\geq c$ , and note that
$(abc)^2\leq 3a^3 \implies a\geq\frac{(bc)^2}{3}.$ On the other hand,
$a^2\mid b^3+c^3\implies a^2\leq 2b^3.$ Combining both bounds gives $\tfrac{(bc)^4}{9}\leq 2b^3\implies bc^4\leq 18$ . Thus $c=1$ and $b\leq 18$ . Check the remaining cases manually.

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GeoMetrix

924 posts

GeoMetrix

#10 h Sep 23, 2020, 12:35 PM • 4 Y

Y by amar_04, Mango247, cubres, Dowsley

Uhh too ez . Similiar to the others but ill post anyways.

Solution

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grupyorum

1418 posts

grupyorum

#11 h Sep 23, 2020, 1:05 PM • 2 Y

Y by Kontrol, cubres

Using divisibility twice seems to accelerate the solution, as follows.

Let $a\ge b\ge c$ , without loss. Then $a^2b^2c^2\le 3a^2$ implying $b^2c^2\le 3a$ . Furthermore, $a^2\mid b^3+c^3$ then gives $b^3+c^3\ge a^2$ . Combining these, $b^3+c^3\ge a^2 \ge \frac{b^4c^4}{9}$ . If $c\ge 2$ , then $2b^3\ge b^3+c^3 \ge 16b^4/9$ , giving $b\le 9/8$ , not possible. Thus $c=1$ .

Now we arrive at $a^3+b^3+1=a^2 b^2$ . Next, it is easily seen that $a\ne b$ , thus $a^3+b^3+1< 2a^3$ . Consequently, $b^2<2a$ . Furthermore, $a^2 \mid b^3+1$ . Combining these, we have $b^3+1 \ge a^2 >b^4/4$ . This inequality is possible only when $2\le b\le 4$ . Now, we also have the condition $a^2\mid b^3+1$ , and $a>b$ . Under this, the cases $b=3,4$ turn out to be impossible. Thus $b=2$ . This gives $a=3$ . From symmetry, any permutation of $(1,2,3)$ works.

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pggp

89 posts

pggp

#12 h Sep 23, 2020, 1:05 PM • 2 Y

Y by AFSA, cubres

It's really similar to Iran Second Round problem. It was used on a TST in my country and I knew the main solution idea beforehand.

https://artofproblemsolving.com/community/c6h1832202p12271047

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fukano_2

492 posts

fukano_2

#13 h Sep 24, 2020, 11:44 AM • 1 Y

Y by cubres

WLOG $a \ge b \ge c.$ We have $(abc)^2=a^3+b^3+c^3 \le 3a^3 \Longrightarrow (bc)^2 \le 3a.$
But since $a^2 \le b^3+c^3 \Longrightarrow (bc)^2 \le 3\sqrt{b^3+c^3}$ $\Longrightarrow (bc)^4 \le 9(b^3+c^3),$ which fails if both $b, c \ge 3.$
Bash out all possibilities to get $(b,c) = (2,3) \Longrightarrow (a,b,c) = (1,2,3)$

This post has been edited 2 times. Last edited by fukano_2, Sep 24, 2020, 11:52 AM

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OlympusHero

17020 posts

OlympusHero

#15 h Sep 24, 2020, 5:16 PM • 1 Y

Y by cubres

anyone else notice this? shortlist didn't come out in 2018

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franzliszt

23531 posts

franzliszt

#16 h Sep 24, 2020, 5:21 PM • 1 Y

Y by cubres

OlympusHero wrote:

anyone else notice this? shortlist didn't come out in 2018

See post #4

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coolmath_2018

2807 posts

coolmath_2018

#50 h Feb 19, 2023, 4:37 AM • 1 Y

Y by cubres

WLOG $a \ge b \ge c$ . Note that $a^2 | (b^3 + c^3)$ because we can divide by $a^2$ in the original equation and since the RHS is an integer the LHS must be an integer too. So we know that $a^2 \le b^3 + c^3$ . By our assumption, $3a^3 \ge a^3 + b^3 + c^3 = (abc)^2$ . Hence we know that $a \ge \frac{(bc)^2}{3}$ or $a^2 \ge \frac{b^4c^4}{9}$ . This means that $\frac{b^4c^4}{9} \le b^3 + c^3 \Rightarrow bc^4 \le 18.$ Hence we know that $c= 1$ . This now means that $b \le 18$ . We can test every value and we quickly see that any permutation of $(1,2,3)$ work.

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everythingpi3141592

85 posts

everythingpi3141592

#51 h Feb 25, 2023, 3:42 AM • 2 Y

Y by LostDreams, cubres

By the AM-GM inequality, $abc$ is at most $\left(\frac{a+b+c}{3}\right)^3$ . So rearranging terms in the inequality, $3^6 \geq \frac{\left(a+b+c\right)^3}{a^3+b^3+c^3}\left(a+b+c\right)^3$ . This implies $a+b+c \leq 8$

Now, we can manually check that we cannot have a solution where $2$ of the $3$ numbers are equal, so we must have $(1, 2, 3)$ and $(1, 2, 4)$ and $(1, 3, 4)$ as the only possible solutions out of which only $(1, 2, 3)$ works

This post has been edited 2 times. Last edited by everythingpi3141592, Feb 25, 2023, 3:45 AM

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bobthegod78

2982 posts

bobthegod78

#52 h Jun 22, 2023, 8:29 PM • 1 Y

Y by cubres

Funky solution.

Without loss of generality, let $a\geq b \geq c$ . It follows that $a^2b^2c^2 \leq 3a^3$ , or $b^2c^2 \leq 3a$ . Since $a^2 \mid b^3+c^3$ , we also have $a^2 \leq b^3+c^3$ . From the first inequality, $\frac 19 b^4c^4 \leq a^2$ , so
$\frac 19 b^4c^4 \leq b^3+c^3 \implies 1\leq \frac 9{bc^4} + \frac 9{b^4c}$ If $c\geq 2$ , it is easy to check there are no solutions to this inequality. Then $c=1$ , and we can have any value of $b$ between $1$ and $9$ inclusive. Checking all of them, we find only $(3,2,1)$ works up to permutation.

This post has been edited 2 times. Last edited by bobthegod78, Jun 22, 2023, 8:32 PM

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bjump

1026 posts

bjump

#54 h Jul 6, 2023, 1:07 AM • 1 Y

Y by cubres

solution

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minusonetwelth

225 posts

minusonetwelth

#56 h Oct 31, 2023, 9:11 PM • 1 Y

Y by cubres

The answer is $(a,b,c)=(1,2,3)$ and permutations.

As the equation is symmetric, we may assume that $a\ge b\ge c$ . First note that $a^2\mid b^3+c^3$ , so $a^2\le b^3+c^3.$ Moreover,
$3a^3\ge a^2+b^2+c^2=a^2b^2c^3,$ so $3a\ge b^2c^2$ or $a^2\ge\frac{b^4c^4}{9}$ . Combining these two inequalities,
$\frac{b^4c^4}{9}\le b^3+c^3\implies c^4b\le 9\left(1+\frac{b^3}{c^3}\right)\le 18,$ giving $c=1$ . Therefore, the equation reduces to
$a^3+b^3+1=a^2b^2.$ Then, as $a^2\le b^3+1$ , we have
$a^3+a^2\ge a^3+b^3+1=a^2b^2\implies a+1\ge b^2\implies a^2\ge (b^2-1)^2.$ Putting this together, we have $b^3+1\ge(b^2-1)^2$ , which is inductively false for $b>2$ . So $b=1$ or $b=2$ . Finishing, we see that $b=1$ yields no solution while $b=2$ gives $a=3$ .

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HamstPan38825

8863 posts

HamstPan38825

#57 h Dec 22, 2023, 3:34 AM • 1 Y

Y by cubres

It's like they made ``annoying size details" into a problem :/

Let $a \leq b \leq c$ . Notice that $c^2 \mid a^3+b^3$ and cyclic permutations, which implies that $c \leq \sqrt{a^3+b^3}$ . On the other hand, $3c^3 \geq a^3+b^3+c^3 = a^2b^2c^2 \implies c \geq \frac{a^2b^2}3.$ Now assume $a, b \geq 2$ , such that $ab \geq a+b$ . Then $\frac{a^4b^4}9 \geq \frac{(a+b)^4}9 > \frac{a^4+4a^3b+4ab^3+b^4}9 > a^3+b^3$ as $a+4b \geq 10 > 9$ and $4a+ b>9$ .

So we may assume $a = 1$ , from where the given equation becomes $\frac{b^4}9 < b^3+1$ . We need to check $b \in \{1, 2, \dots, 9\}$ , which yields the only solution triple $(1, 2, 3)$ and cyclic permutations.

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Thapakazi

61 posts

Thapakazi

#58 h Mar 10, 2024, 6:12 PM • 1 Y

Y by cubres

WLOG let, $a \geq b \geq c$ . We have $a^2 \mid b^3 + c^3 \implies a^2 \leq b^3 + c^3.$ On the other hand we have $3a^3 \geq (abc)^2 \implies a \geq \frac{(bc)^2}{3}.$ Putting all of them together, we see

$\frac{(bc)^4}{9} \leq b^3 + c^3 \implies \frac{bc^4}{9} \leq 1 + \left(\frac{c}{b}\right)^3 \leq 2.$
Hence, $bc^4 \leq 18$ which gives $c = 1$ . Thus our equation becomes $a^2 + b^2 + 1 = (ab)^3.$ Note that $a \neq 1$ because $(1,1,1)$ is not a solution. Hence $a > 1.$

Proceeding similarly, we have $2a^3 \geq (ab)^2 - 1 \implies 2a \geq {b^2} - \frac{1}{a^2} > b^2 - 1.$

This implies $a \geq \frac{b^2}{2}.$ But as, $a^2 \leq b^3 + 1 \implies \frac{b^4}{4} \leq b^3 + 1.$

This gives us that $b \leq 4.$ Checking each cases manually we find that only $b = 2$ works. This gives us $a = 3.$ Hence the solutions are $(3,2,1)$ and permutations.

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cursed_tangent1434

626 posts

cursed_tangent1434

#59 h Jun 16, 2024, 11:33 AM • 2 Y

Y by ihategeo_1969, cubres

Solved with ihategeo_1969. We found it funny that this is actually the base case of 2019 Iran MO Round 2 Problem 3 but I guess there's probably a perfectly valid explanation for this. We claim that the only triple of solutions (upto permutations) is $(1,2,3)$ . The equations is entirely symmetric so we assume $a\ge b\ge c$ .

Now, since $a^2$ is a factor of the left hand side it must also be a factor of the right hand side, so $a^2\mid b^3+c^3$ . Then, $a^2\le b^3+c^3$ . But, note that,
$3a^3 \ge a^3+b^3+c^3 = (abc)^2$ So, $3a \ge b^2c^2$ . Combining this with our previous inequality we have that if $c>1$ ,
$18c^3 \ge 9(b^3+c^3)\ge 9a^2 \ge (b^2c^2)^2 = b^4c^4 \ge 16c^4 \ge 32c^3$ which is a very clear contradiction since $c$ is a positive integer. Thus, $c=1$ . Repeating the bounding with this additional piece of information,
$18=18c^3 \ge 9(b^3+c^3)\ge 9a^2 \ge (b^2c^2)^2 = b^4c^4 = b^4$ so $b=1$ or $b=2$ . Of these, when $b=1$ we have $a^2=a^3+2$ which quite clearly has no solutions, and when $b=2$ we have $4a^2 = a^3+9$ which implies $a\le 4$ of which only $a=3$ work giving us our desired solution set.

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Ywgh1

139 posts

Ywgh1

#60 h Aug 17, 2024, 7:13 PM • 1 Y

Y by cubres

2019 N2

We claim that the answer is $(3,2,1)$ upto permutations.
Assume WLOG $a \geq b \geq c$ , then we have that $3a \ge b^2c^2$ , and $a^2 | b^3+c^3$ .
Combining these we get that $c=1$ .

Now our equation is.
$a^3+b^3+1=(ab)^2$ This implies that $a^2 | b^3+1 = (b+1)(b^2+1-b)$ .
But we have that $a^2 > b+1$ and $a^2 \geq b^2+1-b$ , where equality is when $b=a=1$ , hence $a^2 > b^2+1-b$ , implying that $a^2=b^3+1$ , subtracting one we get that.
$(a-1)(a+1)= b^3$
For the other case, if $a | b+1$ and $a| b^2+1-b$ this implies that $a=b+1$ and that $b+1 | b^2+1-b.$
Which holds when $a=3$ and $b=2$ . Hence we are done

This post has been edited 2 times. Last edited by Ywgh1, Aug 18, 2024, 6:30 AM

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AshAuktober

1005 posts

AshAuktober

#62 h Dec 21, 2024, 2:13 PM • 1 Y

Y by cubres

WLOG $a \le b \le c$ . Note that $(abc)^2 \le 3c^3 \implies a^2b^2 \le 3c$ , and $c^2 \mid a^3+b^3 \implies 9(a^3+b^3) \ge a^4+b^4$ .
Analysing $f(a, b) = a^4+b^4 - 9 (a^3+b^3)$ , we see it is increasing in both $a, b \in \mathbb{N}$ and positive for $(2, 2)$ , so in fact $a = 1$ .
From here similar arguments yield $b^4 \le 4b^3+4 \implies b \le 4$ , and case bashing gives $(a, b, c) = (1, 2, 3)$ and permutations.

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asdf334

7585 posts

asdf334

#63 h Dec 22, 2024, 9:01 PM • 2 Y

Y by megarnie, cubres

nice

Only two observations are needed. Assume $a\le b\le c$ . Then we get:

First, notice $(abc)^2=a^3+b^3+c^3\le 3c^3\implies (ab)^2\le 3c$ .
Second, notice $a^3+b^3\equiv a^3+b^3+c^3=(abc)^2\equiv 0\pmod{c^2}$ .

Hence write $k:=\frac{a^3+b^3}{c^2}$ .

Notice that
$a^3+b^3\le 1^3+(ab)^3\le 1+(3c)^{3/2}\implies kc^2\le 1+(3c)^{3/2}\implies k\le \frac{1}{c^2}+\sqrt{\frac{27}{c}}.$ From here we simply bound: notice that $c\ge 28$ fails, as we obtain
$1\le k\le \frac{1}{28^2}+\sqrt{\frac{27}{28}}\implies \frac{28^2-1}{28^2}\le \sqrt{\frac{27}{28}}\implies \left(\frac{28^2-1}{28^2}\right)^2\le \frac{27}{28}\implies 29\cdot 27\cdot 29\le 28\cdot 28\cdot 28$ or if $x:=28$ then $x^3+x^2-x-1=(x-1)(x+1)^2\le x^3$ which is false. Thus $1\le c\le 27$ .

Notice now that $a^3+b^3\le 1+(3c)^{3/2}\le 730\implies 1\le b\le 9$ . We split into cases:

If $b=1$ , then $a=1$ ---and thus $c^2\mid 2$ , so that $a=b=c=1$ , which fails.
If $b=2$ , then $a=1$ or $a=2$ . In the first case, we find $c^2\mid 9$ , so that $c=3$ . The solution $(1,2,3)$ works. In the second case, we find $c^2\mid 16$ , so that $c=2$ or $c=4$ , neither of which work.
If $b=3$ , then $a=1$ , $a=2$ , or $a=3$ , so that $a^3+b^3\in \{28,35,54\}$ . In this case, only $c=3$ is valid, from $c^2\mid 54$ . This fails the original equation.
If $b=4$ , then $a\in \{1,2,3,4\}$ , so that $a^3+b^3\in \{65,72,91,128\}$ . These numbers yield $c=6$ from $c^2\mid 72$ and $c\in {4,8}$ from $c^2\mid 128$ . None of these work---the check, by the way, is being done by checking if $a^3+b^3+c^3$ is a square.
If $b=5$ , then $a\in \{1,2,3,4,5\}$ , so that $a^3+b^3\in \{126,133,152,189,250\}$ . Only $c=5$ works from $c^2\mid 250$ , and this fails.
If $b=6$ , then $a\in \{1,2,3,4,5,6\}$ , so that $a^3+b^3\in \{217,224,243,280,341,432\}$ . Only $c=9$ works from $c^2\mid 243$ , as well as $c\in \{6,12\}$ from $c^2\mid 432$ . None of these work.
If $b=7$ , then $a\in \{1,2,3,4,5,6,7\}$ , so that $a^3+b^3\in \{344,351,370,407,468,559,686\}$ . Only $c=7$ works from $c^2\mid 686$ , and this fails.
If $b=8$ , then $a\in \{1,2,3,4,5,6\}$ from $a^3+b^3\le 730$ , so that $a^3+b^3\in \{513,520,539,576,637,728\}$ . From here, only $c\in\{8,12,24\}$ work from $c^2\mid 576$ . These each fail the original equation.
If $b=9$ , then $a=1$ from $a^3+b^3\le 730$ , and no value works for $c^2\mid 730$ .

We have exhausted all cases. The only working triplet is $(1,2,3)$ and its permutations.

This post has been edited 2 times. Last edited by asdf334, Dec 23, 2024, 3:45 PM

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Assassino9931

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Assassino9931

#64 h Feb 12, 2025, 8:17 PM • 1 Y

Y by cubres

Lol, see China TST 1987.

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N3bula

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N3bula

#65 h Mar 3, 2025, 8:50 AM

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Suppose $c\geq b\geq a$ and that $a^3+b^3\geq c^3$ , we get that $ab\geq ^3\sqrt{c^3-1}$ , thus $(abc)^6\geq (c^3-1)^2c^6$ , we can't have that $(abc)^6\geq (c^3-1)^2c^6> 27c^9$ as $3c^3\geq a^3+b^3+c^3$ , however we have that $(c-1)c^5\geq(c^3-1)^2$ , thus $(c-1)c^11\leq 27c^9$ , thus $(c-1)c^2\leq 27$ , however if $c\geq 4$ this is not possible. Thus now suppose $c\geq 4$ , we have that $a^3+b^3+c^3\leq 2c^3$ , thus now suppose $a^3+b^3\geq c^2$ , if $a^3+b^3<c^2$ we that there are no solutions, as if $ab>c$ we get that $(abc)^2\geq c^3+c^2$ , so thus $ab\leq c$ , however this is not possible as $a^3+b^3\geq 1$ . Thus let $a^3+b^3\geq c^2$ , we get $ab\geq ^3\sqrt{c^2-1}$ , thus $(abc)^6\geq (c^2-1)^2c^6$ , we must have that $(c^2-1)^2c^6 \leq 8c^9$ , however this gives: $c^10-2c^8+c^6\leq 8c^9$ , however for $c\geq 9$ this does not hold, thus $c\leq 8$ . After checking the small cases we find the only solution is any permutation of $(1, 2, 3)$ .

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Frd_19_Hsnzde

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Frd_19_Hsnzde

#66 h Mar 19, 2025, 12:21 AM

Y by

This problem is so smashed but i am posting anyways

.

Since this equation is symmetric then WLOG $a \ge b \ge c$ .

$3a^3 \ge a^3 + b^3 + c^3 = (abc)^2$ this means $3a \ge (bc)^2$ .

And we know that $a^2 \vert a^3 + b^3 + c^3$ i.e. $a^2 \vert b^3 + c^3$ from this $a^2 \le b^3+c^3$ .

Finally we can say that $3\sqrt{b^3+c^3} \ge (bc)^2$ i.e. $9(b^3+c^3) \ge (bc)^4$ .This inequality is wrong when $b \ge 3$ and $c \ge 2$ both happens.Then $b \le 2$ and $c = 1$ . Then after the checking cases there is a unique answer $(3 , 2 , 1)$ and their permutations.

This post has been edited 2 times. Last edited by Frd_19_Hsnzde, Mar 19, 2025, 12:30 AM

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Ilikeminecraft

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Ilikeminecraft

#67 h Apr 26, 2025, 5:28 AM

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Note that $a^2\mid b^3 + c^3.$ Let $a \leq b \leq c.$ Hence, $a^2b^2c^2 \leq 3a^3,$ and thus $b^2c^2 \leq 3a.$ By earlier, we have that $\frac{b^4c^4}{9}\leq a^2 \leq b^3 + c^3 \leq 2c^3.$ Hence, $b^4 c \leq 18.$ Clearly, $b \leq 2.$

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