1979 USAMO Problems/Problem 1


Determine all non-negative integral solutions $(n_1,n_2,\dots , n_{14})$ if any, apart from permutations, of the Diophantine Equation $n_1^4+n_2^4+\cdots +n_{14}^4=1599$.

Solution 1

Recall that $n_i^4\equiv 0,1\bmod{16}$ for all integers $n_i$. Thus the sum we have is anything from 0 to 14 modulo 16. But $1599\equiv 15\bmod{16}$, and thus there are no integral solutions to the given Diophantine equation.

Solution 2

In base $16$, this equation would look like: \[n_1^4+n_2^4+\cdots +n_{14}^4=63F_{16}\]

We notice that the unit digits of the LHS of this equation should equal to $F_{16}$. In base $16$, the only unit digits of fourth powers are $0$ and $1$. Thus, the maximum of these $14$ terms is 14 $1's$ or $E_{16}$. Since $E_{16}$ is less than $F_{16}$, there are no integral solutions for this equation.

See Also

1979 USAMO (ProblemsResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5
All USAMO Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS