Prove that every integer greater than 7 can be written by using 3's and 5's only. that is, for every n > 7 there exist non-negative integers tn, sn such that n = 3tn + 5sn.

An integer may be a multiple of 3.
An integer may be 1 greater than a multiple of 3.
An integer may be 2 greater than a multiple of 3.

It is redundant to say an integer is 3 greater than a multiple of 3 (that's just a multiple of 3, we've got it covered).  Same for 4, 5, 6, 7...

Let's consider a number which is a multiple of 3.  Clearly, we can write 3+3+3+3+... until we reach the number.  It can be written as only 3's.

Let's consider a number which is 2 greater than a multiple of 3.  If we subtract 5 from that number, it'll be a multiple of 3.  That means we can write the number as 5+3+3+3+3+...  Of course, the number must be at least 8.

Let's consider a number which is 1 greater than a multiple of 3.  If we subtract 5 from that number, it'll be 2 greater than a multiple of 3.  If we subtract another 5, it'll be a multiple of 3.  That means we can write the number as 5+5+3+3+3+3+...  Of course, the number must be at least 13.

That's it.  We considered all the numbers.  We forgot 9, 10, 11, and 12, but these are easy peasy.

Beautiful question.