a.) A 7-tuple $(a_1,a_2,a_3,a_4,b_1,b_2,b_3)$ of pairwise distinct positive integers with no common factor is called a shy tuple if $$ a_1^2+a_2^2+a_3^2+a_4^2=b_1^2+b_2^2+b_3^2$$and for all $1 \le i<j \le 4$ and $1 \le k \le 3$, $a_i^2+a_j^2 \not= b_k^2$. Prove that there exists infinitely many shy tuples. b.) Show that $2016$ can be written as a sum of squares of four distinct natural numbers.
Problem
Source: RMO Delhi, 2016 P5
Tags: number theory, Sum of Squares
11.10.2016 10:16
Take $a_1=x-3, a_2=x-1, a_3=x+1, a_a=x+3, b_1=2x, b_2=2, b_3=4$ for any even $x\ge 4$. Noting that $a_i$'s are odd and $b_j$'s are even, it is trivial to show that this gives us a shy tuple for infinite values of $x$. Done! I admit this seems like unintutive magic. I could think of this because of this identity: $(a+b+c)^2 + (-a+b+c)^2 + (a-b+c)^2 + (a+b-c)^2 = (2a)^2 + (2b)^2 + (2c)^2$ Anyone with a more down-to earth solution?
11.10.2016 10:19
$2016 = 16^2 + 20^2 + 24^2 + 28^2$ No guess... No trial... No brute force... No computer program... Just some intuition and the use of the above identity along with some calculation.
11.10.2016 10:28
$b)2016=40^2+16^2+12^2+4^2=36^2+20^2+16^2+8^2=32^2+28^2+12^2+8^2=28^2+24^2+20^2+16^2$
11.10.2016 10:34
Note that $a_{1}, a_{2}, a_{3}, a_{4} = 4k+1, 3k+2, 2k+4, k+3$ and $b_{1}, b_{2}, b_{3} = 5k+2, 2k+5, k+1$ also works. We just need to note that $1^2+2^2+3^2+4^2=5^2+2^2+1^2$
11.10.2016 15:09
$(k-1)^2 + (k+1)^2 +49^2 = (k-7)^2 + (k+7)^2 + 1 + 48^2$ With $k = 30(MOD42)$
11.10.2016 21:08
guessing 2016 one . worth how many pts..?
12.10.2016 07:53
Interestingly equation in this form. http://www.artofproblemsolving.com/community/c3046h1046714__
12.10.2016 11:34
What do they mean by 'no common factor'?
12.10.2016 20:15
http://www.artofproblemsolving.com/community/c3046h1318437_4_squares_with_3_squares
12.10.2016 21:28
What if we take the values as $ k, k+1, k+2, k+3, 2k+3, 1, 2 $ We can check that it works for $ k>2 $ I did it this way. But what does $pairwise$ distinct with no common factor mean? Doesn't it mean that any two numbers have to be coprime and not taking the gcd of whole 7-tuple as 1 ? Someone clarify what does the question statement say? Rishabh_Ranjan wrote: Take $a_1=x-3, a_2=x-1, a_3=x+1, a_a=x+3, b_1=2x, b_2=2, b_3=4$ for any even $x\ge 4$. Noting that $a_i$'s are odd and $b_j$'s are even, it is trivial to show that this gives us a shy tuple for infinite values of $x$. Done! I admit this seems like unintutive magic. I could think of this because of this identity: $(a+b+c)^2 + (-a+b+c)^2 + (a-b+c)^2 + (a+b-c)^2 = (2a)^2 + (2b)^2 + (2c)^2$ Anyone with a more down-to earth solution? Even this has to be wrong solution in that case. Statement confusion :/
12.10.2016 21:31
Btw, by my solution we can get the second part's answer with ease. The same as - Rishabh_Ranjan wrote: $2016 = 16^2 + 20^2 + 24^2 + 28^2$ No guess... No trial... No brute force... No computer program... Just some intuition and the use of the above identity along with some calculation. It's just that I'm a bit confused by the statement
31.01.2017 11:50
roopsa wrote: What do they mean by 'no common factor'? They actually want to mean gcd(a,b)=1 when two distinct numbers a,b is choosen from that tuple. Is it help? Rather All guys and friends, so called brothers and sisters here solve this problem in pretty constructive manner. Well Second one Can easily be follows from Lagrange Four Square Identity. For The First One I have Different Approach. Although Solutions here are great.
31.01.2017 13:30
@sauditya can you please detail the solution and prove rigrously that theyvare all pairwise co prime also this should do it (k+2)^2 +(k-2)^2+9^2+20^2=(k-8)^2+(k+8)^2 +19^2. Also using chinese remainder theorem assures that there exists such a k.
31.01.2017 18:41
is m y solution correct?
30.05.2017 22:22
the_executioner wrote: Note that $a_{1}, a_{2}, a_{3}, a_{4} = 4k+1, 3k+2, 2k+4, k+3$ and $b_{1}, b_{2}, b_{3} = 5k+2, 2k+5, k+1$ also works. We just need to note that $1^2+2^2+3^2+4^2=5^2+2^2+1^2$ No it will not work if $k$ is odd then then $k+3$ and $k+1$ is even and if $k$ is even then $3k+2$ and $5k+2$ is even So?? Rather use Some Different Concept to do this I did some what Like this many days before, by Creating maps and maps and maps to that famous right angle triangle triplet. and finally use the face that $p=4k+1=a^2+b^2$ are solvable in $p,a,b$ if $p$ is prime by Pigeonhole you can prove that. $(a-b)^{2}+(c-d)^{2}=a^{2}+b^{2}+c^{2}+d^{2}-2 (ab+cd)$ now add $(ab+cd)^2+1$ on both side we have, $(a-b)^{2}+(c-d)^{2}+(ab+cd)^{2}+1=(ab+cd-1)^{2}+a^{2}+b^{2}+c^{2}+d^{2} $ Now , substitute $a=u^{2}-v^{2}$ and $b=2uv $ $c=m^{2}-n^{2} $ and $d=2mn $ then finally apply $a^{2}+b^{2}=(u^{2}+v^{2})^2$ similarly to $c $ and $d $ to get $(a-b)^{2}+(c-d)^{2}+(ad+cd)^{2}+1=(ab+cd-1)^{2}+(u^{2}+v^{2})^{2}+(m^{2}+n^{2})^2$ Finally with $p=u^{2}+v^{2}=(4r+1)$ and $q=m^{2}+n^{2}=(4t+1) $ and $p$ and $q $ distinct primes. This ends the problem.
28.05.2018 14:14
Most probably the question means that the 7 numbers do not have any common factor (but does not imply that they are pairwise coprime). I couldn't find the official solution to confirm this. If this is the case, can someone confirm if my solution is correct. Please. My solution: Take $(a_1,a_2,a_3,a_4,b_1,b_2,b_3)=(m^2+n^2,12k,9k,36k, m^2-n^2, 2mn, 39k)$, where $m$ is odd, $n$ is even and $k$ is any integer that makes each of the 7 numbers distinct. There are infinitely many such $m,n,k$, so infinitely many such 7-tuples.
28.05.2018 14:20
I would like to ask that the question paper of rmo for delhi and other places are different???
28.05.2018 15:04
It was different in 2016, but same in 2017.
28.05.2018 15:17
Oh but why was it different in 2016
28.05.2018 15:29
Smita wrote: Oh but why was it different in 2016 2016 wasn't the only year in which it was different, up until 2017 RMO used to be different papers for 3-5 regions. In 2017, the number of regions was cut down to only 2. The reason wasn't really specified.
28.05.2018 18:32
Severus wrote: My solution: Take $(a_1,a_2,a_3,a_4,b_1,b_2,b_3)=(m^2+n^2,12k,9k,36k, m^2-n^2, 2mn, 39k)$, where $m$ is odd, $n$ is even and $k$ is any integer that makes each of the 7 numbers distinct. There are infinitely many such $m,n,k$, so infinitely many such 7-tuples. Please tell me if this correct.
29.05.2018 16:25
Bumping this, pleez halp!!
29.05.2018 16:44
(a) Consider the identity $(x+y+z)^2 + (x+y-z)^2 + (x-y+z)^2 + (-x+y+z)^2 = (2x)^2 + (2y)^2 + (2z)^2$. This is satisfied for all real values $x, y$ and $z$, and hence, it is satisfied for any positive integer as well. (b) $(2x)^2 + (2y)^2 + (2z)^2 = 2016 \Rightarrow 4x^2 + 4y^2 + 4z^2 = 2016 \Rightarrow x^2 + y^2 + z^2 = 504$ Now, Legendre's three-square theorem states that: Every number that is not of the form $4^m(8k+7)$ is the sum of three integer squares. $504 = 4\cdot124$, but $124 \equiv 0$ ($mod$ $8$). Hence, 2016 can be written as the sum of three integer squares, and using the original equation, 2016 can also be written as the sum of four integer squares.
30.07.2018 21:42
Rishabh_Ranjan wrote: Take $a_1=x-3, a_2=x-1, a_3=x+1, a_a=x+3, b_1=2x, b_2=2, b_3=4$ for any even $x\ge 4$. Noting that $a_i$'s are odd and $b_j$'s are even, it is trivial to show that this gives us a shy tuple for infinite values of $x$. Done! I admit this seems like unintutive magic. I could think of this because of this identity: $(a+b+c)^2 + (-a+b+c)^2 + (a-b+c)^2 + (a+b-c)^2 = (2a)^2 + (2b)^2 + (2c)^2$ Anyone with a more down-to earth solution? If I am not making mistake, if you consider the word "no common factors" as, the all seven has no common factor (instead of they "pairwise don't have common factor ") then you don't need to say "they are pairwise distinct". So, I think we need $7$-tuple don't have common factor pairwise. Please correct me if I am wrong.
31.12.2018 18:51
SAUDITYA wrote: $(k-1)^2 + (k+1)^2 +49^2 = (k-7)^2 + (k+7)^2 + 1 + 48^2$ With $k = 30(MOD42)$ How did you come up with it? Can you share the motivation behind it?
28.01.2021 02:10
Considering the word "with No common factor" As all 7 seven together have No common factor. So We can choose tuples such that $a_1^2+a_2^2+a_3^2=b_1^2$ and $b_2^2 +b_3^2=a_4^2$ as both have infinitely many solution. So we can get infinitely many tuples.
24.02.2022 10:13
$(a)$ Taking $a_{1} = 6, a_{2} = 2, a_{3} = 3$ and $b_{1} = 7$ gives $6^2 + 2^2 + 3^2 + a_{4}^2 = 7^2 +b _{2}^2 + b_{3}^2$ and since we know $a_{4}^2 = b _{2}^2 + b_{3}^2$ has infinitely many solutions we have proved the required and have adhered to the given conditions. $(b)$ Through some intuition we have, $2016 = 24^2 + 32^2 + 20^2 + 4^2$
18.10.2024 08:10
This works right? $6^2+2^2+3^2+(5k)^2=(4k)^2+(3k)^2+7^2.$