1. ## Divisibly Problems

Here are 3 problems that i couldn't solve them. I would wonder if anybody help me.

a) If $5|3a+4b$ prove that $25|2a^2-3ab+2b^2$
b) If $a^9|b^5$ prove that $a^7|b^4$
c) If $a, b, c$ are natural numbers and $a^b|b^a$ and $b^c|c^b$, prove that $a^c|c^a$

2. Originally Posted by havaliza
Here are 3 problems that i couldn't solve them. I would wonder if anybody help me.

a) If $5|3a+4b$ prove that $25|2a^2-3ab+2b^2$
b) If $a^9|b^5$ prove that $a^7|b^4$
c) If $a, b, c$ are natural numbers and $a^b|b^a$ and $b^c|c^b$, prove that $a^c|c^a$
a) is false. For example, if a=2 and b=1 then 3a+4b = 10 (multiple of 5), but 2a^2-3ab+2b^2 = 4 (not a multiple of 25).

b) Suppose that p is one of the prime factors of a. Then p must also be a divisor of b. If $p^s$ is the highest power of p that divides a, and $p^t$ is the highest power of p that divides b, then the condition $a^9|b^5$ implies that $9s\leqslant5t$, or $\tfrac st\leqslant\tfrac59$. Conversely, if that condition holds for each prime divisor of a, then $a^9|b^5$. The result then follows from the fact that $\tfrac59<\tfrac47$.

c) Hint: notice that $a^{bc} | b^{ac} | c^{ba}$.

3. Thanks for the help. The first problem was right but i had mistyped it

a) If $5|3a+4b$ prove that $25|2a^2-3ab-2b^2$

4. Originally Posted by havaliza
Thanks for the help. The first problem was right but i had mistyped it

a) If $5|3a+4b$ prove that $25|2a^2-3ab-2b^2$
In that case, notice that $2a^2-3ab-2b^2 = (a+2b)(2a-b)$, and show that each of those factors can be expressed as a multiple of 3a+4b plus some multiple of 5.