Consider modulo 7, the left side is always divisibile by 7, whereas, the right side is divisible by 7 for no integer b.

$LHS \equiv 0 \pmod{7}$ $7 \not \vert RHS$.

Since 7 consecutive numbers appear on left side, it's a good idea to try 7 remainders. $1^3+2^3+3^3+4^3+5^3+6^3+7^3\equiv 1+1+(-1)+1+(-1)+(-1)+0\equiv 0 (mod7)$, so left side is always divisible by 7. Right side mod 7 is not 0, look at consecutive numbers that are on the 4th power, and add up their remainders. Does RMO stand for Romanian Mathematical Olympiad?

@Above, no, It stands for Regional Mathematics Olympiads, the first (or second?) stage for selection of IMO team in India

No, regional mathematical olympiad, Indian.

Putting a=x-3 in the equation gives LHS=7x^3+14*6x Hence 7 divides LHS.And it's easy to see 7 doesn't divide RHS.Hence no solution

Hello Guys. One of my batchmates solved the problem like this: He factorized the LHS and claimed that it is composite and then claimed that RHS is prime for every integer b ( I don't know how he did that ). I want to know, is this correct? I tried putting integers from 0 to 5 and found that RHS was always prime but I could not find any way to prove it for all integers b. Is RHS , in this case, always prime? could someone please show me how?

TheLegendOfLegends wrote: I want to know, is this correct? no.

RHS=$b^4+(b+1)^4=p(b) $ is a polynomial in the variable $b$. By Goldbach we know that there cannot exist any polynomial that gives primes as the output for all values of $b$

So, will he receive any marks for his solution?

TheLegendOfLegends wrote: Hello Guys. One of my batchmates solved the problem like this: He factorized the LHS and claimed that it is composite and then claimed that RHS is prime for every integer b ( I don't know how he did that ). I want to know, is this correct? I tried putting integers from 0 to 5 and found that RHS was always prime but I could not find any way to prove it for all integers b. Is RHS , in this case, always prime? could someone please show me how? Of course RHS is not prime for all $b$. He will get zero. For example, $f(1) = 17$. So $f(1+17k) \equiv f(1) \equiv 0 \pmod {17}$ for all $ k \geq 1$. But $f(1+17k)$ is large, so it is definitely not prime.

I would like to share how I solved the problems: 1. I took a point P, made an arbitrary angle AOB with scale, drew a line parallel to OA ( using compass and ruler ) through P and made it intersect with OB at X. Then I bisected OX and used that bisected portion as radius to cut OA at C such that XC=1/2 XO. Then I joined PC and extended it to meet OB at D. By BPT, I proved the ratio 1:2 2. I assumed that Integers satisfy the equation. I added all terms in the LHS to get the form 7k, k is an integer. On the RHS, I added all terms to get the form 2m+1, m is an integer. After that I got confused as to how I should proceed. So I made 2 cases, when a is an odd integer and when a is an even integer. I proved for the first case that for any integer b and any odd integer a, equation has no solutions because 7k was even while 2m + 1 is odd.. But, then I could not prove the same for second case and lost any hopes to get the solution ( I did not study Modular Arithmetic so I could not use the mod 7 thing) 3. I got to the final equation but I did a calculation mistake and wrote the constant term as -4, while it was -2. So I could not get any roots for the equation. 4. Left it blank 5. Simply drew the diagram and bid farewell to the question. 6. I did not know how to solve it so I did this: I took the LHS to the RHS . Then I had the terms greater than equal to 0. The three terms were of the form 2(x-y)/y^2 - 1 . Then I made all terms denominators same and proceeded only to see the time ran out. I know I messed up. Could someone speculate how many marks I can get out 102?

TheLegendOfLegends wrote: I would like to share how I solved the problems: 1. I took a point P, made an arbitrary angle AOB with scale, drew a line parallel to OA ( using compass and ruler ) through P and made it intersect with OB at X. Then I bisected OX and used that bisected portion as radius to cut OA at C such that XC=1/2 XO. Then I joined PC and extended it to meet OB at D. By BPT, I proved the ratio 1:2 2. I assumed that Integers satisfy the equation. I added all terms in the LHS to get the form 7k, k is an integer. On the RHS, I added all terms to get the form 2m+1, m is an integer. After that I got confused as to how I should proceed. So I made 2 cases, when a is an odd integer and when a is an even integer. I proved for the first case that for any integer b and any odd integer a, equation has no solutions because 7k was even while 2m + 1 is odd.. But, then I could not prove the same for second case and lost any hopes to get the solution ( I did not study Modular Arithmetic so I could not use the mod 7 thing) 3. I got to the final equation but I did a calculation mistake and wrote the constant term as -4, while it was -2. So I could not get any roots for the equation. 4. Left it blank 5. Simply drew the diagram and bid farewell to the question. 6. I did not know how to solve it so I did this: I took the LHS to the RHS . Then I had the terms greater than equal to 0. The three terms were of the form 2(x-y)/y^2 - 1 . Then I made all terms denominators same and proceeded only to see the time ran out. I know I messed up. Could someone speculate how many marks I can get out 102? You should rather post it in the post-RMO discussion thread in the India forums.

Ok. Thanks for the advice.

biomathematics wrote: TheLegendOfLegends wrote: Hello Guys. One of my batchmates solved the problem like this: He factorized the LHS and claimed that it is composite and then claimed that RHS is prime for every integer b ( I don't know how he did that ). I want to know, is this correct? I tried putting integers from 0 to 5 and found that RHS was always prime but I could not find any way to prove it for all integers b. Is RHS , in this case, always prime? could someone please show me how? Of course RHS is not prime for all $b$. He will get zero. For example, $f(1) = 17$. So $f(1+17k) \equiv f(1) \equiv 0 \pmod {17}$ for all $ k \geq 1$. But $f(1+17k)$ is large, so it is definitely not prime. Thank you for your solution. But, just one thing, if f(1+17k) is large, then how can you say it is definitely not prime?

TheLegendOfLegends wrote: biomathematics wrote: TheLegendOfLegends wrote: Hello Guys. One of my batchmates solved the problem like this: He factorized the LHS and claimed that it is composite and then claimed that RHS is prime for every integer b ( I don't know how he did that ). I want to know, is this correct? I tried putting integers from 0 to 5 and found that RHS was always prime but I could not find any way to prove it for all integers b. Is RHS , in this case, always prime? could someone please show me how? Of course RHS is not prime for all $b$. He will get zero. For example, $f(1) = 17$. So $f(1+17k) \equiv f(1) \equiv 0 \pmod {17}$ for all $ k \geq 1$. But $f(1+17k)$ is large, so it is definitely not prime. Thank you for your solution. But, just one thing, if f(1+17k) is large, then how can you say it is definitely not prime? It is greater than 17 and is divisible by 17, so not prime

Say $b=2k$,$b^4+(b+1)^4=32k^4+24k^2+8k+32k^3. \equiv 0(mod7)$ cannot occur bcoz residues possible are $1,3,–1,.$

$ $

.......... Redacted

Mr.Techworm wrote: Consider modulo 7, the left side is always divisibile by 7, whereas, the right side is divisible by 7 for no integer b. Can you pls tell why RHS is not divisible by 7

$x^4\equiv 0,1,2,4\pmod{4}$ hence if sum of two fourth power is divisible by $7$ then both of them are divisible by $7$.but $\gcd(a, a+1)=1$ so it's impossible

Let's assume $x=a+3$ . $\implies (a^3+(a+1)^3+\cdots+(a+6)^3) =\{(x-3)^3+(x+3)^3\}+\{(x+2)^3+(x-2)^2\} +...+x^3$. $=7a^3+12a=7a(a^2+12)$. So if the given equation has any solution then $7|b^4+(b+1)^4$. Clearly for $b=7k$ it is not possible . For $b=7k+1$, $b^4+(b+1)^4 \equiv 3 \pmod{7}$. For $b=7k+2$ ,$b^4+(b+1)^4 \neq 0\pmod{7}$. For $b=7k+3$ also $b^4+(b+1)^4\neq 0\pmod{7}$. Note that $4,5,6$ are can be represent as $-3,-2-1$ modulo 7.so $4^4+5^4+6^4$ will as same as $3^4+2^4+1$ respectively modulo 7. So we get contradiction !!!

DeathlyHallow wrote: Mr.Techworm wrote: Consider modulo 7, the left side is always divisibile by 7, whereas, the right side is divisible by 7 for no integer b. Can you pls tell why RHS is not divisible by 7 Easy.... x^4 is congruent to 0, 1, 2, 4 (mod 7) Notice that the sum of any two remainders will not reach 0 or 7

Just the easiest problem in rmo 2017

Using the closed form for the sum of cubes we get $$\sum_{j=0}^{6}(a+j)^3 =\sum_{k=1}^{a+6} k^3 -\sum_{k=1}^{a-1} k^3 =\left(\frac{(a+6)(a+7)}{2} \right)^2 -\left(\frac{a(a+1)}{2} \right)^2 =7\cdot \left(\frac{2a+6}{4} \right) \equiv 0\,(mod\, 7)$$Then we've a contradiction by noting that the sum of any two distinct biquadratic residues $(mod\, 7)$ is always $\not\equiv 0\,(mod\,7)$ Biquadratic residues modulo 7

$\blacksquare$

Notice $\{a,a+1,\dots a+6\}\equiv\{0,1,\dots,6\}\pmod{7}$ so $a^3+(a+1)^3+\dots+(a+6)^3\equiv 7\cdot 8/2\equiv 0\pmod{7}.$ On the other hand, $0,1,4,2$ are the QRs modulo $7$ so $b^4+(b+1)^4\equiv1,5,6,2\not\equiv 0\pmod{7}.$ $\square$

oi bruv the sum of two perfect squares will only be divisible by 7 if both numbers are divisible by 7 innit. and it aint hard to see that the rhs is divisible by 7 ( once you factorise it and ting) hence proved yeah

FTSOC, we assume integer solutions exist.Clearly, the left-hand side is divisible by $7$. So, for integer solutions $(a,b)$ to exist, we must have that $7|b^4+(b+1)^4$ which is not possible. This is a contradiction, and we are done!

$\pmod{7}$ works.