Finding the smallest positive integer a

Question

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

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 06:32

score 5 · Accepted Answer · answered Jul 31 '11 at 06:31

5

HINT:

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

answered Jul 31 '11 at 06:31

davidlowryduda

91,687

My answer is 512.I used Diophantine equations to minimize a.Thank you for the suggestion. – Eisen Jul 31 '11 at 07:02

score 3 · Answer 2 · answered Jul 31 '11 at 06:36

3

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

answered Jul 31 '11 at 06:36

Jyrki Lahtonen

133,153

Nicer idea than the solution in Engel's book. – André Nicolas Jul 31 '11 at 06:56
@Andre: I don't have the book. What is the solution? – davidlowryduda Jul 31 '11 at 07:20
@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

score 1 · Answer 3 · edited Apr 13 '17 at 12:20

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.

score 0 · Answer 4 · edited Mar 27 '13 at 08:08

0

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

edited Mar 27 '13 at 08:08

Dominic Michaelis

19,935

answered Mar 27 '13 at 07:45

user68694

1

How is this relevant to the question? – robjohn Mar 27 '13 at 09:19

score 0 · Answer 5 · edited Apr 13 '17 at 12:20

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, $$ a\equiv-50\cdot857\equiv512\pmod{1971} $$

Finding the smallest positive integer a

5 Answers5