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?

$2016 = 16^2 + 20^2 + 24^2 + 28^2$ No guess... No trial... No brute force... No computer program... Just some intuition and the use of the above identity along with some calculation.

$b)2016=40^2+16^2+12^2+4^2=36^2+20^2+16^2+8^2=32^2+28^2+12^2+8^2=28^2+24^2+20^2+16^2$

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 $1^2+2^2+3^2+4^2=5^2+2^2+1^2$

$(k-1)^2 + (k+1)^2 +49^2 = (k-7)^2 + (k+7)^2 + 1 + 48^2$ With $k = 30(MOD42)$

guessing 2016 one . worth how many pts..?

Interestingly equation in this form. http://www.artofproblemsolving.com/community/c3046h1046714__

What do they mean by 'no common factor'?

http://www.artofproblemsolving.com/community/c3046h1318437_4_squares_with_3_squares

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 :/

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

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.

@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.

is m y solution correct?

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.

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.

I would like to ask that the question paper of rmo for delhi and other places are different???

It was different in 2016, but same in 2017.

Oh but why was it different in 2016

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.

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.

Bumping this, pleez halp!!

(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.

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.

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?

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.

$(a)$ 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, $2016 = 24^2 + 32^2 + 20^2 + 4^2$

This works right? $6^2+2^2+3^2+(5k)^2=(4k)^2+(3k)^2+7^2.$