Question 1203198
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3443 = 11 * 313


This shows 3443 is not prime. It is composite.
It doesn't matter which base you are working with because we can convert between any two bases.
A prime number in one base, is a prime number in any base.


Some examples:
{{{3443[5] = (21[5])*(2223[5])}}}
{{{3443[7] = (14[7])*(625[7])}}}
{{{3443[4] = (23[4])*(10321[4])}}}
The subscript represents which base we're working in
For instance {{{3443[5] = matrix(1,3,3443,"base",5)}}}


Further Reading:
<a href="https://math.stackexchange.com/questions/3999/is-a-prime-number-still-a-prime-when-in-a-different-base">https://math.stackexchange.com/questions/3999/is-a-prime-number-still-a-prime-when-in-a-different-base</a>
and
<a href="http://web.archive.org/web/20190714164706/http://mathforum.org/library/drmath/view/55880.html">http://web.archive.org/web/20190714164706/http://mathforum.org/library/drmath/view/55880.html</a>


Useful calculator
<a href="https://www.rapidtables.com/convert/number/base-converter.html">https://www.rapidtables.com/convert/number/base-converter.html</a>


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Another approach


{{{3443[b] = 3b^3 + 4b^2 + 4b + 3}}}
Use the rational root theorem to determine that b = -1 is a root of {{{3b^3 + 4b^2 + 4b + 3}}}
Therefore, (b+1) is a factor


Use polynomial long division, or the shortcut synthetic division, to find that {{{(3b^3 + 4b^2 + 4b + 3)/(b+1) = 3b^2+b+3}}}


That rearranges to {{{3b^3 + 4b^2 + 4b + 3 = (b+1)(3b^2+b+3)}}}
For {{{(b+1)(3b^2+b+3)}}} to be prime, one of the factors must be 1.
If b+1 = 1, then b = 0. But we can't have base 0.


If 3b^2+b+3 = 1, then it leads to two nonreal complex roots.
The base cannot be complex as only positive integers are allowed
Specifically from the set {5,6,7,8,9,...} so we can form {{{3443[b]}}}


We conclude that neither factor (b+1) nor (3b^2+b+3) can be 1.
Therefore, {{{3b^3 + 4b^2 + 4b + 3 = (b+1)(3b^2+b+3)}}} is never prime. It is always composite.
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