Find the last two digits of the number $7^{100}-8^{100}$. So I tried and found that $$7^{100} \equiv 1 \pmod{100}$$ but I got stuck with $8^{100}$. Help me out please. 
 A: First, note that $7^2\equiv8^2\pmod{5}$. So $$7^{100}-8^{100}=(7^2-8^2)\left((7^2)^{49}+(7^2)^{48}(8^2)+\cdots+(8^2)^{49}\right)\equiv 0\pmod{25},$$
since the second term consists of $50$ summands of the same modulo $5$. 
Second $7^{100}-8^{100}\equiv 7^{100}\pmod{4}\equiv 1\pmod{4}$.
Chinese remainder theorem then implies that the answer is $25$.
A: Consider the binomial expansion
$$(2+5)^{100}=2^{100}+100\cdot2^{99}\cdot5+{100\choose2}2^{98}\cdot5^2+\cdots+{100\choose98}2^2\cdot5^{98}+100\cdot2\cdot5^{99}+5^{100}$$
It's easy to see that all the interior terms are divisible by $2^2\cdot5^2=100$, hence
$$7^{100}=(2+5)^{100}\equiv2^{100}+5^{100}\equiv2^{100}+25\mod100$$
It's even easier to see that expanding the binomial for $(2-10)^{100}$ gives
$$8^{100}=(2-10)^{100}\equiv2^{100}\mod100$$
Hence
$$7^{100}-8^{100}\equiv25\mod100$$
(Remark:  I tried $7=10-3$ first, before discovering that $2+5$ did the trick.)
A: Following David Cipras advice let me give this answer:
The conjugate rule applied twice gives us 
$7^{100}-8^{100}= (7^{25}-8^{25})(7^{25}+8^{25})(7^{50}+8^{50})$ 
The first factor is 1-1 mod (25) = 0 mod (25) so the last numbers are 00,-25 or -75 mod(100) 
And considering mod 3 we have:
$7^{100}-8^{100} = 1^{100}-(-1){100}=1-1=0$ mod(3) so the only alternatives are 00 or -75 mod (100).
I now must also consider mod (2) (not mentioned in my comment) to get 
$7^{100}-8^{100}= 1-0 $ mod (2) which rules out the alternative 00 and gives us -75 mod (100) that is the last two numbers are "25".
PS I am still worried about the negative numbers as mentioned in my first comment. Actually wolfram alpha gives  -2037032741857976461510454343761609060834657536462737346010739515459049908071877069255337375 
Perhaps there is some kind of convention about this? 
