sqing wrote: Find the smallest positive integer $m$ satisfying the following condition: for all prime number $p$ such that $p\ge3$,have $105|9^{ p^2}-29^p+m.$ (September 28, 2012, Hohhot) Are u sure that $p\geq 3$ ? because for $p=3$ we get $7|m$ and for $p>3,7|m+1$ so for all prime there doesn't exist no $m$. If it was given $p>3$ then clearly $5|m,21|m+1$ so $m\geq 20$

subham1729 wrote: sqing wrote: Find the smallest positive integer $m$ satisfying the following condition: for all prime number $p$ such that $p\ge3$,have $105|9^{ p^2}-29^p+m.$ (September 28, 2012, Hohhot) Are u sure that $p\geq 3$ ? because for $p=3$ we get $7|m$ and for $p>3,7|m+1$ so for all prime there doesn't exist no $m$. If it was given $p>3$ then clearly $5|m,21|m+1$ so $m\geq 20$ $p>3 .$ I'm sorry. Thanks.

$m$ can be found by solving the following system of modulo $m\equiv 2 \pmod 3$, $m\equiv 0 \pmod 5$ and $m \equiv 6 \pmod 7$ (you want to formally use CRT) edit: got $m=20$

The above solutions are wrong. The real solution is $m=55$ which comes from solving the system $m\equiv 1 \pmod 3$ $m\equiv 0 \pmod 5$ $m\equiv 6 \pmod 7$ Which give $m\equiv 55 \pmod {105}$ and hence our minimal solution.

m=20 is the right answer, the solution above is definitely wrong.

Recall that for all primes $p>3$, we have $p^2 \equiv 1$ (mod $24$). Taking mod $105$ hereinafter, we firstly have $29^p \equiv 29 \cdot 841^{\frac{p-1}{2}} \equiv 29$. Note that $36^a \equiv 36$ for any positive integer $a$, we also have $9^{p^2} = 9 \cdot (9^{24})^k \equiv 9 \cdot ((-6)^8)^k \equiv 9 \cdot 36^k \equiv 9$, where we write $k= \frac{p^2-1}{24}$. Note that $9-29+m \equiv 0 \implies m \equiv 20$. Hence the desired value of $m$ is $20$.