# How can I prove that $10=2^{a}*3^{b}*7^{c}$ has infinite solutions?

Both in a unrescrited case and with the following restriction: $$a+b+c=1$$

For the unrestricted case, let $$a=1.$$ Now you need $$3^b7^c=5$$, so for a positive integer $$n$$, let

$$3^b = 5^{n+1}, 7^c = 5^{-n}$$.

Solve these two equations for $$b$$ and $$c$$.

That should give you a leg up on the restricted case.

Hint $$c=1-a-b$$

Your equation then becomes $$\frac{10}{7}=(\frac{2}{7})^a (\frac{3}{7})^b$$

Show now that if you fix $$a$$ you can find some $$b$$ which satisfies this equation.