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Can we find the smallest positive integer $a$ such that $1971|50^n+a.23^n$ where n is odd?

Source:Problem Solving Strategies by Arthur Engel

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Presumably you want this to hold for all odd values of $n$? Is that period between $a$ and $23^n$ a multiplication sign? The TeX-code \cdot gives you that. – Jyrki Lahtonen Jul 31 '11 at 6:32
I added an IMHO relevant tag. – Jyrki Lahtonen Jul 31 '11 at 6:42
up vote 5 down vote accepted


1971 = 27 * 73. Use modular arithmetic and congruences.

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My answer is 512.I used Diophantine equations to minimize a.Thank you for the suggestion. – Eisen Jul 31 '11 at 7:02

Hint: $50^2\equiv 23^2\pmod{1971}$

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Nicer idea than the solution in Engel's book. – André Nicolas Jul 31 '11 at 6:56
@Andre: I don't have the book. What is the solution? – mixedmath Jul 31 '11 at 7:20
@mixedmath: $50^n+23^n a\equiv (-4)^n +(-4)^na\equiv -4^n(a+1) \pmod{27}$, $50^n+23^n a \equiv (-23)^n +23^n a \equiv 23^n(a-1)\pmod{73}$. Then standard solving of $a\equiv -1\pmod{27}$, $a\equiv 1\pmod{73}$. – André Nicolas Jul 31 '11 at 12:41

For coprime $\rm\: b,c\in \mathbb Z\:,\ \ a\: =\: -(b/c)^{2\:k+1} =\: -(b^2/c^2)^{k}\ b/c\: \equiv\: -b/c \pmod{b^2-c^2}\:.\:$
So the extended Euclidean algorithm will efficiently compute $\rm\:a \equiv -b/c\pmod{b^2-c^2}\:.$

Alternatively note $\rm\:a\equiv -1\pmod{b-c}$ since then $\rm\ b\equiv c\ \Rightarrow\ a = -b/c \equiv -c/c\equiv -1\:.\: $
Similarly we infer $\rm\ a\:\equiv\ 1\ \pmod{b+c}\:.\:$ When $\rm\:b,c\:$ have opposite parity, $\rm\:b-c,\ b+c\:$ are coprime, so we may employ $\rm CRT$ to efficiently compute the unique solution $\rm\: (mod\ \ b^2-c^2)\:.$

Such nontrivial $(\ne \pm 1)$ square-roots of $1\:$ exist modulo composite $\rm\:m\:$ that are not prime powers. In fact, given such a nontrivial square root $\rm\:a\:$ one may compute a factor of $\rm\:m\:$ by $\rm\:gcd(a\pm1,m)\:,\:$ e.g. above $\rm\ a = 512,\ \ gcd(511,1971) = 73,\ \ gcd(513,1971) = 27\:.\:$ This is the way many integer factoring algorithms work, e.g. Fermat's method of difference of squares and its generalizations, e.g. MPQS. See here for more on relations between factorization, nontrivial sqrts and idempotents.

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Find the smallest positive integer $n$ for: $$\left(\frac{1+j}{1-j}\right)^n =1;\quad (j^2 =-1)$$

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How is this relevant to the question? – robjohn Mar 27 '13 at 9:19

Since $50^2\equiv23^2\equiv529\pmod{1971}$ and $(529,1971)=1$, we have $$ \begin{align} 50^{2n+1}+a\cdot23^{2n+1}&\equiv0\pmod{1971}\\ 50\cdot529^n+a\cdot23\cdot529^n&\equiv0\pmod{1971}\\ 50+a\cdot23&\equiv0\pmod{1971} \end{align} $$ Using the Euclid-Wallis Algorithm $$ \begin{array}{r} &&85&1&2&3&2\\\hline 1&0&1&-1&3&-10&23\\ 0&1&-85&86&-257&857&-1971\\ 1971&23&16&7&2&1&0\\ \end{array} $$ we get that $857\cdot23\equiv1\pmod{1971}$. Therefore, $$ a\equiv-50\cdot857\equiv512\pmod{1971} $$

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