Anik's remainders

Anik likes division and is always eager to find remainders.

Once Anik divided a x a^x by ( a 1 ) (a-1) to get positive remainder r 1 r_1 ; and by ( a + 1 ) (a+1) to get positive remainder r 2 r_2 .

If Anik assumed a a as a positive integer greater than 2 and x x as an positive odd number greater than 1, find r 1 + r 2 r_1+r_2 .

Try more of my original problems in this set (click here) .
I made this problem as a simple generalization of this problem .
a x ax a x a^x a 1 a-1 a + 1 a+1 a a a x + 1 ax+1

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1 solution

Nihar Mahajan
Jun 9, 2015

We crack this problem using simple congruence:

a 1 ( m o d a 1 ) a x 1 ( m o d a 1 ) r 1 = 1 a \equiv 1 \pmod{a-1} \\ \Rightarrow a^x \equiv 1 \pmod{a-1} \\ \Rightarrow \boxed{r_1=1}

Now we will use the fact that x x is odd now:

a 1 ( m o d a + 1 ) a x 1 a ( m o d a + 1 ) r 2 = a a \equiv -1 \pmod{a+1} \\ \Rightarrow a^x \equiv -1 \equiv a \pmod{a+1} \\ \Rightarrow \boxed{r_2=a}

Thus , r 1 + r 2 = a + 1 \huge\boxed{r_1+r_2=a+1} .

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