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Difference between revisions of "1982 USAMO Problems/Problem 4"
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== Solution ==
 
== Solution ==
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{{solution}}
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Indeed, <math>\boxed{k=2935363331541925531}</math> has the requisite property.
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To see why, consider the primes <math>3,\ 5,\ 17,\ 257,\ 65537,\ 6700417,\ 641</math>, and observe that <cmath>\begin{align*}k&\equiv 1\pmod{3}\\ k&\equiv 1\pmod{5}\\ k&\equiv 1\pmod{17}\\ k&\equiv 1\pmod{257}\\ k&\equiv 1\pmod{65537}\\ k&\equiv 1\pmod{6700417}\\ k&\equiv -1\pmod{641}\end{align*}</cmath>
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Moreover, <cmath>\begin{align*}\text{ord}_{3}\left(2\right)&=2\\ \text{ord}_{5}\left(2\right)&=4\\ \text{ord}_{17}\left(2\right)&=8\\ \text{ord}_{257}\left(2\right)&=16\\ \text{ord}_{65537}\left(2\right)&=32\\ \text{ord}_{6700417}\left(2\right)&=64\\ \text{ord}_{641}\left(2\right)&=64\end{align*}</cmath>
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We conclude that <cmath>\begin{align*}n\equiv 1\pmod{2}&\implies k\cdot 2^{n}-1\equiv k\cdot\left(-1\right)+1\equiv 0\pmod{3}\\ n\equiv 2\pmod{4}&\implies k\cdot 2^{n}-1\equiv k\cdot\left(-1\right)+1\equiv 0\pmod{5}\\ n\equiv 4\pmod{8}&\implies k\cdot 2^{n}-1\equiv k\cdot\left(-1\right)+1\equiv 0\pmod{17}\\ n\equiv 8\pmod{16}&\implies k\cdot 2^{n}-1\equiv k\cdot\left(-1\right)+1\equiv 0\pmod{257}\\ n\equiv 16\pmod{32}&\implies k\cdot 2^{n}-1\equiv k\cdot\left(-1\right)+1\equiv 0\pmod{65537}\\ n\equiv 32\pmod{64}&\implies k\cdot 2^{n}-1\equiv k\cdot\left(-1\right)+1\equiv 0\pmod{6700417}\\ n\equiv 0\pmod{64}&\implies k\cdot 2^{n}-1\equiv k\cdot1+1\equiv 0\pmod{641}\end{align*}</cmath>
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And <math>k>>3,5,17,257,65537,6700417,641</math> so the relevant values will, in fact, always be composite. <math>\blacksquare</math>
  
 
== See Also ==
 
== See Also ==

Latest revision as of 00:15, 4 June 2021

Problem

Prove that there exists a positive integer $k$ such that $k\cdot2^n+1$ is composite for every integer $n$.

Solution

Indeed, $\boxed{k=2935363331541925531}$ has the requisite property.

To see why, consider the primes $3,\ 5,\ 17,\ 257,\ 65537,\ 6700417,\ 641$, and observe that \begin{align*}k&\equiv 1\pmod{3}\\ k&\equiv 1\pmod{5}\\ k&\equiv 1\pmod{17}\\ k&\equiv 1\pmod{257}\\ k&\equiv 1\pmod{65537}\\ k&\equiv 1\pmod{6700417}\\ k&\equiv -1\pmod{641}\end{align*}

Moreover, \begin{align*}\text{ord}_{3}\left(2\right)&=2\\ \text{ord}_{5}\left(2\right)&=4\\ \text{ord}_{17}\left(2\right)&=8\\ \text{ord}_{257}\left(2\right)&=16\\ \text{ord}_{65537}\left(2\right)&=32\\ \text{ord}_{6700417}\left(2\right)&=64\\ \text{ord}_{641}\left(2\right)&=64\end{align*}

We conclude that \begin{align*}n\equiv 1\pmod{2}&\implies k\cdot 2^{n}-1\equiv k\cdot\left(-1\right)+1\equiv 0\pmod{3}\\ n\equiv 2\pmod{4}&\implies k\cdot 2^{n}-1\equiv k\cdot\left(-1\right)+1\equiv 0\pmod{5}\\ n\equiv 4\pmod{8}&\implies k\cdot 2^{n}-1\equiv k\cdot\left(-1\right)+1\equiv 0\pmod{17}\\ n\equiv 8\pmod{16}&\implies k\cdot 2^{n}-1\equiv k\cdot\left(-1\right)+1\equiv 0\pmod{257}\\ n\equiv 16\pmod{32}&\implies k\cdot 2^{n}-1\equiv k\cdot\left(-1\right)+1\equiv 0\pmod{65537}\\ n\equiv 32\pmod{64}&\implies k\cdot 2^{n}-1\equiv k\cdot\left(-1\right)+1\equiv 0\pmod{6700417}\\ n\equiv 0\pmod{64}&\implies k\cdot 2^{n}-1\equiv k\cdot1+1\equiv 0\pmod{641}\end{align*}

And $k>>3,5,17,257,65537,6700417,641$ so the relevant values will, in fact, always be composite. $\blacksquare$

See Also

1982 USAMO (Problems • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

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