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Difference between revisions of "1982 USAMO Problems/Problem 4"
(→‎Solution: Essentially chooses Fermat Numbers until We get to a composite one, 1+1/2+...+1/32+1/64+1/64=1, will write motivation soon)
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Prove that there exists a positive integer <math>k</math> such that <math>k\cdot2^n+1</math> is composite for every integer <math>n</math>.
 
Prove that there exists a positive integer <math>k</math> such that <math>k\cdot2^n+1</math> is composite for every integer <math>n</math>.
  
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== Solution ==
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== Solution 1 ==
  
 
Let <math>p</math> be a prime number that divides <math>2^y-1</math> and <math>x</math> be a whole number less than <math>y</math> such that <cmath>k\equiv -1\cdot2^{y-x}\pmod{p}</cmath> If <math>n-x</math> is a multiple of <math>y</math>, then, for some integer <math>r</math>, <cmath>k\cdot2^{n}\equiv-1\cdot2^{y-x}\cdot2^{x+ry}\pmod{p}</cmath> This simplifies to <cmath>k\cdot2^{n} \equiv -1\pmod{p}</cmath> This implies that <math>k\cdot2^{n}+1\equiv 0 \pmod{p}</math>. Thus we conclude that there exists an integer <math>k</math> such that <math>k\cdot2^{n}+1</math> is composite for all integral <math>n</math>.
 
Let <math>p</math> be a prime number that divides <math>2^y-1</math> and <math>x</math> be a whole number less than <math>y</math> such that <cmath>k\equiv -1\cdot2^{y-x}\pmod{p}</cmath> If <math>n-x</math> is a multiple of <math>y</math>, then, for some integer <math>r</math>, <cmath>k\cdot2^{n}\equiv-1\cdot2^{y-x}\cdot2^{x+ry}\pmod{p}</cmath> This simplifies to <cmath>k\cdot2^{n} \equiv -1\pmod{p}</cmath> This implies that <math>k\cdot2^{n}+1\equiv 0 \pmod{p}</math>. Thus we conclude that there exists an integer <math>k</math> such that <math>k\cdot2^{n}+1</math> is composite for all integral <math>n</math>.
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== Solution 2==
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I claim that <math>\boxed{k=2935363331541925531}</math> works
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Consider the primes <math>3,5,7,17,257,65537,6700417,641</math>
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Note that <cmath>k\equiv 1\text{ (mod }3,5,7,17,257,65537,6700417)</cmath> and that <cmath>k\equiv 1\text{ (mod }641)</cmath>
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Also, <cmath>\text{ord}_3(2)=2</cmath><cmath>\text{ord}_5(2)=4</cmath><cmath>\text{ord}_{17}(2)=8</cmath><cmath>\text{ord}_{257}(2)=16</cmath><cmath>\text{ord}_{65537}(2)=32</cmath><cmath>\text{ord}_{6700417}(2)=2=\text{ord}_{641}(2)=64</cmath>
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Take <math>m</math> to be an odd integer.
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It is well known (and not hard to prove) that <math>2^{\frac{\text{ord}_p(2)}{2}\cdot m}\equiv \left(2^{\frac{\text{ord}_p(2)}{2}}\right)^m\equiv -1^m\equiv -1\text{ (mod }p)</math>
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Consider some cases:
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When <math>n\equiv1\text{ (mod }2)\iff n=m\ </math> we have <math>2^n\cdot k+1\equiv -k+1\equiv 0 \text{ (mod }3)</math>
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When <math>n\equiv2\text{ (mod }4)\iff n=2m\ </math> we have <math>2^n\cdot k+1\equiv -k+1\equiv 0 \text{ (mod }5)</math>
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When <math>n\equiv4\text{ (mod }8)\iff n=4m\ </math> we have <math>2^n\cdot k+1\equiv -k+1\equiv 0 \text{ (mod }17)</math>
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When <math>n\equiv8\text{ (mod }16)\iff n=8m\ </math> we have <math>2^n\cdot k+1\equiv -k+1\equiv 0 \text{ (mod }257)</math>
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When <math>n\equiv16\text{ (mod }32)\iff n=16m\ </math> we have <math>2^n\cdot k+1\equiv -k+1\equiv 0 \text{ (mod }65537)</math>
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When <math>n\equiv32\text{ (mod }64)\iff n=32m\ </math> we have <math>2^n\cdot k+1\equiv -k+1\equiv 0 \text{ (mod }6700417)</math>
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When <math>n\equiv0\text{ (mod }64)\iff64\mid n\ </math> we have <math>2^n\cdot k+1\equiv k+1\equiv 0 \text{ (mod }641)</math>, since <math>2^{64}\equiv 1  \text{ (mod }641)</math>
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And furthermore, <math>k>>3,5,17,257,65537,6700417,641</math> so these numbers need be composite.
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But this covers all cases; we are done <math>\Box</math>
  
 
== See Also ==
 
== See Also ==

Revision as of 22:00, 21 September 2016

Problem

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

Solution 1

Let $p$ be a prime number that divides $2^y-1$ and $x$ be a whole number less than $y$ such that \[k\equiv -1\cdot2^{y-x}\pmod{p}\] If $n-x$ is a multiple of $y$, then, for some integer $r$, \[k\cdot2^{n}\equiv-1\cdot2^{y-x}\cdot2^{x+ry}\pmod{p}\] This simplifies to \[k\cdot2^{n} \equiv -1\pmod{p}\] This implies that $k\cdot2^{n}+1\equiv 0 \pmod{p}$. Thus we conclude that there exists an integer $k$ such that $k\cdot2^{n}+1$ is composite for all integral $n$.

Solution 2

I claim that $\boxed{k=2935363331541925531}$ works

Consider the primes $3,5,7,17,257,65537,6700417,641$

Note that \[k\equiv 1\text{ (mod }3,5,7,17,257,65537,6700417)\] and that \[k\equiv 1\text{ (mod }641)\]

Also, \[\text{ord}_3(2)=2\]\[\text{ord}_5(2)=4\]\[\text{ord}_{17}(2)=8\]\[\text{ord}_{257}(2)=16\]\[\text{ord}_{65537}(2)=32\]\[\text{ord}_{6700417}(2)=2=\text{ord}_{641}(2)=64\]


Take $m$ to be an odd integer.

It is well known (and not hard to prove) that $2^{\frac{\text{ord}_p(2)}{2}\cdot m}\equiv \left(2^{\frac{\text{ord}_p(2)}{2}}\right)^m\equiv -1^m\equiv -1\text{ (mod }p)$

Consider some cases:

When $n\equiv1\text{ (mod }2)\iff n=m$ we have $2^n\cdot k+1\equiv -k+1\equiv 0 \text{ (mod }3)$

When $n\equiv2\text{ (mod }4)\iff n=2m$ we have $2^n\cdot k+1\equiv -k+1\equiv 0 \text{ (mod }5)$

When $n\equiv4\text{ (mod }8)\iff n=4m$ we have $2^n\cdot k+1\equiv -k+1\equiv 0 \text{ (mod }17)$

When $n\equiv8\text{ (mod }16)\iff n=8m$ we have $2^n\cdot k+1\equiv -k+1\equiv 0 \text{ (mod }257)$

When $n\equiv16\text{ (mod }32)\iff n=16m$ we have $2^n\cdot k+1\equiv -k+1\equiv 0 \text{ (mod }65537)$

When $n\equiv32\text{ (mod }64)\iff n=32m$ we have $2^n\cdot k+1\equiv -k+1\equiv 0 \text{ (mod }6700417)$

When $n\equiv0\text{ (mod }64)\iff64\mid n$ we have $2^n\cdot k+1\equiv k+1\equiv 0 \text{ (mod }641)$, since $2^{64}\equiv 1 \text{ (mod }641)$

And furthermore, $k>>3,5,17,257,65537,6700417,641$ so these numbers need be composite.

But this covers all cases; we are done $\Box$

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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