Which of the following statements are true Either give a pro

Which of the following statements are true? Either give a proof or a counterexample! For all a, n sum N \\ {0}, if a|n^2 and a lessthanorequalto n, then a|n. For all a, b sum N \\ {0}, if a|b, then a^2|b^2. For all a, b sum N \\ {0}, if a|13b, then a|13 or a|b. For all primes p and a, b sum N \\ {0} and x, y sum Z, if p|ax + by, then p|a or p|b.

Solution

1 ) True .

Note: This is called \"proof by contradiction\" and it is a very standard mathematical technique.

If \'a\' does not divide n, then n is the product of one or more primes of the form p(i)
and \'a\' cannot be one of them since we have assumed that \'a\' does not divide n.

So n = [p(1) *...* p(k)] where k => 1

Then n^2 = [(p(1) *...* p(k)] * [p(1) *...* p(k)]

But none of those p(i) are \'a\', so that means that \'a\' does not divide n^2.

But this is a contradiction of the original conditions that \'a\' divides n^2 and \'a\' does not divide n, so the conditions are not possible.

Therefore \'a\' does divide n.

2 ) True .

Since a / b ==>    b = ak for some k.

taking square on boths side we get b^2 = k^2 a^2     ===> b^2 = k_1 b^2     we can replace the k by k_1 for some k_1.

this implies that a^2 / b^2

3 ) Given that a / 13b ===> 13 b = ak -------1)

Also we have gcd( a , 13) =1 because 13 is a prime number.

So we know   1 = am + 13n   , multiplying b on both side we get

                  .   b = ab m + 13bn.

                       b = ab m + ak n = a ( bm+ kn) { by 1)}

                       b = a ( bm+ kn)   ==> a / b  

similarly ,   we can prove that a / 13 .

So this statement is true.

4) Let p | a and p | b. Then there exist integers q_1 and q_2 such that a=pq_1 and b=pq_2.
Hence, for integers x and y,
ax+by=pq_1x+pq_2y=p(q_1x+q_2y)

Since q_1x+q_2y
is an integer, p | (ax+by)