Powerful factors

The problem required the use of the facts that

$$\begin{array}{rcl}{x}^2-y^2& =& (x-y)(x+y)\text{and}\\ x^3+y^3& =& (x+y)(x^2-xy+y^2)\end{array}$$

to find the highest power of 2 and the highest power of 3 which divide $5^{36}-1.$

Alexander Marynovsky from Israel sent in this solution.

$$\begin{array}{rcl}{5}^{36}-1& =& (5^{18}-1)(5^{18}+1)\\ & =& (5^9-1)(5^9+1)(5^6+1)(5^{12}-5^6+1)\\ & =& (5^3-1)(5^6+5^3+1)(5^3+1)(5^6-5^3+1)(5^2+1)(5^4-5^2+1)(5^{12}-5^6+1)\\ & =& (5^3-1)(5^6+5^3+1)(5+1)(5^2-5+1)(5^6-5^3+1)\\ & & (5^2+1)(5^4-5^2+1)(5^{12}-5^6+1)\end{array}$$

Remember what we have got here, I'll use it twice (for 2 and for 3).

Now let's take out the 2's from it.

Because $5^n - 5^k$ is even and therefore $5^n - 5^k +1$ is odd, $(5^2- 5+1), (5^6- 5^3 +1), (5^4 - 5^2 +1), (5^{12} -5^6 +1)$ obviously can't be divided by 2.

So we are left with

$$(5^3-1)(5^6+5^3+1)(5+1)(5^2+1)=124.15751.6.26=2^4.3.13.31.15751$$

So the highest power of 2 is 4.

$$\begin{array}{rcl}(5^2-5+1)(5^6& -& 5^3+1)(5^4-5^2+1)(5^{12}-5^6+1)\\ & =& 21(125.124+1)(25.24+1)(125.125.124.126+1)\\ & =& 21.15501.601(1000.125.31.63+1)\\ & =& 21.15501.601.244125001\end{array}$$

Combining these results:

$$5^{36}-1=2^4.3^3.7.13.31.5167.15751.601.37.6597973$$

The highest power of 3 is 3.

The method can be shortened using modulus arithmetic.