Solve the equation $a^3+b^3+c^3=2001$ in positive integers. Mircea Becheanu, Romania
Problem
Source: JBMO 2001, Problem 1
Tags: geometry, modular arithmetic
30.10.2005 05:36
Don't see the word 'distinct' anywhere... 1,10,10
30.10.2005 05:42
rashad9607 wrote: Don't see the word 'distinct' anywhere... 1,10,10 So what if you don't see the word distinct? Are you sure that is the only solution? Why?
30.10.2005 06:56
09.11.2005 10:57
$\forall t\in\mathbb N t^3\equiv0,\pm1(mod9).$ $2001\equiv3(mod9).$ Hence, $a=3x+1,b=3y+1,c=3z+1$ for $\{x,y,z\}\subset\mathbb N_0.$ $a^3\leq2001.$ Hence, $x\leq3$ and $y\leq3,z\leq3.$ Let $x\geq y\geq z.$ Then $3a^3\geq2001\Rightarrow x>2.$ Hence, $x=3.$ Hence, $b^3+c^3=1001.$ Hence, $2(3y+1)^3\geq1001\Rightarrow y>2.$ Hence, $2<y\leq3.$ Hence, $y=3 \Rightarrow z=0 \Rightarrow a=10,b=10,c=1.$ Well $\{(10,10,1),(10,1,10),(1,10,10)\}.$
03.12.2007 16:02
1234567890 wrote:
can you please tell me isn't this = -1 $ x\equiv5\mod7 \rightarrow x^3\equiv NEGATIVE 1\mod7$ and please elaborate on the conditions for this: So only $ 3^3,6^3,10^3\equiv - 1\mod7$ w/conditions. TQ.
04.12.2007 00:52
If I remember correctly, this problem is like an easier version of N1 IMO shortlist 2002 (it's been a couple months since I saw it though).
04.12.2007 10:47
If we look mod $ 9$, then $ a\equiv b\equiv c\equiv 1\mod 3$. WLOG $ a\ge b\ge c$, then $ 2001 = a^3 + b^3 + c^3\le 3a^3$, additionally $ 2001 = a^3 + b^3 + c^3\ge a^3$, so $ 8.7\le a\le 12.6$. Hence $ a = 10$. Then $ 1001 = b^3 + c^3\le 2b^3$, and $ 1001 = b^3 + c^3\ge b^3$, so $ 10.003 > b > 7.94$ $ b = 10$. Finally, $ c^3 = 1$ so $ (a,b,c) = (10,10,1)$ and permutations.
02.09.2013 04:53
Unfortunately, we only get $a^3, b^3, c^3 \equiv 1 \pmod{9}$ which definitely does not imply $a, b, c \equiv 1 \pmod{9}$; it in fact implies $a, b, c \equiv 1 \pmod{3}$. Also, (not so) surprisingly this is the link from the contests page of Junior Balkan MO 2001. EDIT: unfortunately for me, sunny2000 deleted his post to whom I am replying.
23.05.2019 02:11
I know this ancient but there is sol using mod 4 should I post?
20.11.2020 22:14
TheMathLife wrote: I know this ancient but there is sol using mod 4 should I post? It will be awesome.
20.11.2020 22:26
Valentin Vornicu wrote: Solve the equation $a^3+b^3+c^3=2001$ in positive integers. Mircea Becheanu, Romania as mentioned earlier you can merely use casework as each of a, b, c are 1(mod 3). since there are only 4 vals each can attain the cw isn't too bad
07.12.2020 12:22
Valentin Vornicu wrote: Solve the equation $a^3+b^3+c^3=2001$ in positive integers. Mircea Becheanu, Romania $a^3+b^3+c^3=2001$ as $2001\equiv 3\mod 9$ and $x^3\equiv 0,1,-1\mod 9$ clearly then $a^3,b^3,c^3\equiv 1\mod 9$ then $a,b,c\equiv 1$ or $4$ or $7\mod 9$ and also $a,b,c<13$ Hence we are left with very less cases to try and hence only solution are $\{a, b, c\}\in \{10, 10,1\}$
30.06.2022 14:22
de bash Note that since $2001\equiv 3\pmod 9$, we get that $a^3, b^3, c^3 $ are all $1\pmod 9$, so $a,b,c\equiv 1\pmod3$ and they are all less than $13$. So each of $a,b,c$ are in the set $\{1,4,7,10\}$. Claim: $a,b,c\ne 4$. Proof: Suppose one of them was $4$, let it be $c$. Then $a^3+b^3=2001-64 = 1937 = 13\cdot 149$. So $(a+b)(a^2+ab+b^2) = 13\cdot 149$. Since $13$ and $149$ are primes, and $a^2+ab+b^2>a+b>1$, we have $a+b=13$ and $a^2+ab+b^2 = 149$. There are no positive integer solutions to this. $\square$ Claim: $a,b,c\ne 7$. Proof: Suppose FTSOC $c=7$. Then $a^3+b^3=(a+b)(a^2+ab+b^2)=1658$. Since $1658=2\cdot 829$ and $829$ is prime, we have $a+b=2$ and $a^2+ab+b^2 = 829$, no solutions. $\square$ So each of $a,b,c$ is in $\{1,10\}$. Clearly we must have $\boxed{(10,10,1)}$ or permutations.
07.01.2023 12:40
Shouldn't it be a^3+b^3=(a+b)(a^2-ab+b^2)?
07.10.2023 18:19
Due to symmetry we may assume $a\geq b \geq c$. Then $3a^3 \geq a^3 + b^3 + c^3 = 2001$, i.e. $a^3 \geq 667$, so $a\geq 9$ (as $8^3 = 512$). On the other hand, $a^3 \leq 2001$ implies $a \leq 12$. If $a=9$, then $b^3 + c^3 = 1272$, so $b^3 \geq 636$ and thus $b\geq 9$, i.e. $b=9$. When $b=9$ we get $c^3 = 543$, which has no solution. If $a=10$, then $b^3 + c^3 = 1001$, so $b^3 \geq 501$ and $b\geq 8$. For $b=8$ we obtain $c^3 = 489$, impossible; for $b=9$ we obtain $c^3 = 272$, impossible; and for $b=10$ we obtain $c^3 = 1$, i.e. $c=1$. If $a=11$, then $b^3 + c^3 = 670$, so $b^3 \geq 335$ and $b\geq 7$. For $b=7$ we get $c^3 = 327$, impossible; for $b=8$ we get $c^3 = 158$, while $b^3 < 670$ implies $b \leq 8$. If $a=12$, then $b^3 + c^3 = 273$, so $b^3 \geq 137$ and $b^3 < 273$, implying $b=6$. But for $b=6$ we get $c^3 = 57$, which is impossible. In conclusion, all solutions are $(10,10,1)$, $(10,1,10)$, $(1,10,10)$.
07.10.2023 18:35
Valentin Vornicu wrote: Solve the equation $a^3+b^3+c^3=2001$ in positive integers. Mircea Becheanu, Romania Taking mod $9$, we see that either $a^3$ is equal to $-1,0,1$. Taking mod $9$ yields $a^3+b^3+c^3 \equiv 3 \pmod 9$. We can then see that the values $a,b,c$ must have a remainder of $1$ after being divided by $9$, with the only reasonable values $1$ or $10$. We can then deduce that $(1,10,10)$ and permutations work