a small note about quartic residues mod 16 w/o calculation: if a number is even, clearly the residue is 0 (2^4 = 16). if we have (2k+1)^4 = 16k^4+ 32k^3+16k^2+1 so remainder is 1. edit: oops sorry post below me is right, last terms are 24k^2+8k+1 = 8k^2+8k+1 = 8k(k+1) + 1 and k(k+1) is even.

Not quite ... in fact $(2k+1)^4 = 16k^4 + 32k^3 + 24k^2 + 8k + 1 = 16k^3(k + 2) + 8k(3k+1) + 1$, and since $k(k+3)$ must be even ...

Slightly different proof for the residues: Clearly, if $n$ is even, then $n^4 \equiv 0 \mod 16.$ Now, if $n$ is odd, let $n=2k+1$ for some integer $k\ge 0.$ Thus, \[n^4= (2k+1)^{4}= 16k^{4}+32k^{3}+24k^{2}+8k+1.\] If $k$ is even, then the first four terms of the polynomial are divisible by $16,$ so $n^4 \equiv 1 \mod 16.$ If $k$ is odd, then the first two terms are divisible by $16$ and the third and fourth terms both leave a remainder of $8$ when divided by $16,$ so again, $n^4 \equiv 1 \mod 16.$ Thus, if $n$ is even, $n^4 \equiv 0 \mod 16.$ If $n$ is odd, $n^4 \equiv 1 \mod 16.$

We have $n^4-1=(n^2+1)(n-1)(n+1)$. When $n$ is odd $n^2+1$ is even and so are $n+1$ and $n-1$, but one is divisible by $4$, giving a total of at least $4$ powers of $2$. So When $n$ is odd $n^4\equiv1\pmod{16}$

Easiest USAMO problem ever!

This problem should be in USAJMO P1.

TheMath_boy wrote: This problem should be in USAJMO P1. This is a very old test, so of course the problems will be easier. And JMO didn't exist until 2010 I think.

TheMath_boy wrote: This problem should be in USAJMO P1. Probably even easier.

samrocksnature wrote: TheMath_boy wrote: This problem should be in USAJMO P1. Probably even easier. probably AIME #4-7

Nah that's too hard, I'm able to solve NT up to #10 so probably 12 or 13.

$n^4 \equiv 0, 1 \pmod {16},$ hence there are no solutions modulo 16.

just like 8 is good for squares, 16 is good for 4th powers

Take modulo $16$ gives $-1\pmod{16}$. Now, take fourth power modulo $16$ gives $1, 0, 1, 0$ are the only possible choices, and we need $n_1^4+...+n_{14}^4\equiv -1\pmod{16}$ such that $n_i\in \{0, 1\}~~1\leq i\leq 14$. This is clearly impossible, thus there are no solutions.

Take mod 16, we have $x_1^4 + x_2^4 + \dots + x_{14}^4 \equiv -1 \left( \mod 16 \right)$. This is not possible as the only quartic residues mod 16 are 0 and 1. Thus no solns exist. $\square$

$(2k+1)^4 = 16k^4 + 32k^3 + 24k^2 + 8k + 1 = 16k^3(k + 2) + 8k(3k+1) + 1$, as k is odd then 3k+1=2n, or the RHS is $16k^3(k+2)+16nk+1=1 (mod 16)$. $(2k)^4=0 mod 16$, so the possible values of $x_1^4 + x_2^4 + ... + x_{14}^4$ range from 0 to 14 mod 16, cannot take on 15 mod 16.

is amsp nt level 1 problems similar to this??

Knowing that $n^4$ only leaves remainders $0,1 \pmod{16}$, the sum will be $0 \le x \le 14$, thus it can never equal $15 \pmod{16}$.

Note that 1599 is $15\pmod{16}$, if $n_k$ is even then it is divisible by 16, and if it is odd then it has remainder 1 when divided by 16, if all 14 $k_n$'S are odd then we still have a reminder of 14 which is still not 15.

Use mod $16$ (the quartic residues mod $16$ are $0$ and $1$). Notice that the mod $16$ of the LHS is less than or equal to 14. But the mod $16$ of the RHS is $15$, which is impossible. Therefore, there are no solutions. $\blacksquare$

mod 16 kills prob easiest USAMO problem ever

peelybonehead wrote: mod 16 kills prob easiest USAMO problem ever this is like aime p1 level currently

mathmax12 wrote: peelybonehead wrote: mod 16 kills prob easiest USAMO problem ever this is like aime p1 level currently I can not confirm Apple trees much easier Ok

Still nontrivial if you didn't think mod 16

The only quartic residues modulo 16 are 0 and 1, so $\sum_{i=1}^{14} n_i^4$ can only take on the residues 0 through 14. Since \[1599 \equiv 15~(\text{mod } 16),\]there are $\boxed{\text{no solutions}}$ to the equation.

Clearly all $n_i\le 6$. Notice that $1^4$, $3^4$, and $5^4$ are all $1\pmod{16}$ and $2^4$, $4^4$, and $6^4$ are all $0\pmod{16}$. However, $1599\equiv 15\pmod{16}$. Therefore, there are no solutions.

There are $\boxed{\text{no solutions}}$. Notice in mod $16$ all perfect powers of $4$ are either $0$ or $1$. Hence, it is impossible for the above configuration to occur as $1599 \equiv 15 \pmod{16}$.

Note that $x^4 \equiv \left\{0,1\right\}\pmod{16}$. But we have that $n_1^4 + \cdots + n_{14}^4 \equiv \left\{0,1,\ldots,14\right\}\pmod{16}$. Now, $1599 \equiv 15\pmod{16}$ which gives a contradiction.

Observe that all fourth powers are either $0$ or $1$ mod $16$. Also observe that if the above diophantine has a solution, so does the diophantine $\sum_{i = 1}^{15} n_i^4 = 1600$ with $n_{15} = 1$ . However this is not true as clearly for any solution, $n_i \equiv 0$ mod $16 \forall i$ but $n_{15} = 1$. QED.

mathmax12 wrote: peelybonehead wrote: mod 16 kills prob easiest USAMO problem ever this is like aime p1 level currently To support this claim, this exact problem literally showed up on the problem set of my friends introductory number theory course here at UCSC. So yeah, just goes to show how much easier usamo was when my parents were born.

mho18 wrote: is amsp nt level 1 problems similar to this?? Tbh, this is probably easier than asmp nt level 1

Livesolve from MegaMath Channel

$n^4 \equiv 0, 1 \pmod{16}$ by inspection. But the right side is $15 \pmod{16}$, so there are $\boxed{\text{no solutions}}$