# Solutions to primality testing based on algebra — RMO training

Problem 1:

Given two integers a and m larger than 1, show that, if m is odd, then $n+1$ then $a+1$ is a divisor of $a^{m}+1$. Use this result to obtain the factorization of 1001.

Solution 1:

Since m is odd, we have $a^{m}+1 = (a+1)(a^{m-1}-a^{m-2}+a^{m-3}- \ldots a+1)$ and the result follows.

This shows that

$1001 = 10^{3}+1 = (10+1)(10^{2}-10+1)=11.91=11.7.13$.

Problem 2:

Generalize the result of the above problem to obtain that if a and m are two integers larger than 1 and if $d \geq 1$ is an odd divisor of m, then $a^{m/d}+1$ is a divisor of $a^{m}+1$. Use this result to show that 101 is a factor of 1000001.

Solution 2:

The result is immediate if we write $a^{m}+1=(a^{m/d})^{d}+1$ and then we apply the result of the above problem. This shows that

$1000001 = 10^{6}+1 = (10^{2}+1)(10^{4}-10^{2}+1)=101.9901$.

Problem 3:

Show that 7, 11 and 13 are factors of $10^{15}+1$

Solution 3:

It follows from the previous result that $10^{15}+1 = (10^{3}+1)((10^{3})^{4}-(10^{3})^{3}+(10^{3})^{2}-10^{3}+1)$.

The result then follows from the fact that 7, 11, and 13 factors of 1001.

Problem 4:

Show that $n^{4}+4$ is a composite number for each integer $n \geq 2$. More generally, show that if a is a positive integer such that $2a$ is a perfect square, then $n^{4}+a^{2}$ is a composite number provided that $n \geq \sqrt{2a}$

Solution 4:

It is enough to observe that $n^{4}+4= (n^{2}-2n+2)(n^{2}+2n+2)$. For the general case, we only need to observe that $n^{4}+a^{2}= (n^{2}-\sqrt{2a}n +a)(n^{2}+\sqrt{2a}n+a)$.

We also make an observation that the condition $n \geq \sqrt{2a}$ is sufficient but not necessary.

Cheers,

Nalin Pithwa

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