1979 USAMO Problems/Problem 1

Problem

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.

Video Solution by OmegaLearn

https://youtu.be/zfChnbMGLVQ?t=4778

~ pi_is_3.14

1979 USAMO (Problems • Resources)
Preceded by First Question	Followed by Problem 2
1 • 2 • 3 • 4 • 5
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1979 USAMO Problems/Problem 1

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Problem

Solution 1

Solution 2

Video Solution by OmegaLearn

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