Solution:
\( \displaylines{\text{Given relatively prime integers }p,q\text{ suppose that }\frac{p^2}{q^2}=15.\\ \text{Then }p^2=3\cdot5\cdot q^2\text{ and }3\text{ divides }p.\text{ Let }p=3k.\\ (3k)^2=9k^2=3\cdot5\cdot q^2\Rightarrow3k^2=5q^2.\\ \text{So }3\text{ divides }q,\text{ a contradiction, as }p,q\text{ were relatively prime.}} \)
Explanation:
Note that if 3 does not divide p, it does not divide p^2. Therefore 3 dividing p^2 implies 3 divides p.
Solution:
\( \displaylines{\text{Suppose }\left(\frac{p}{q}\right)^2=15\text{ with }\frac{p}{q}\text{ in lowest terms.}\\ p^2=15q^2\text{ so }} \)
Explanation:
Too difficult to enter in this faux-latex entry system.
Suppose (p/q)^2 = 15 with p/q in lowest terms. Then p^2 = 15q^2, so p^2 is divisible by 3 and 5. Hence p is divisible by 3 and 5, so write p = 3×5×r. Then (15r)^2 = 15 q^2, so 15 r^2 = q^2, and so q^2 is divisible by 3 and 5. Hence q is divisible by 3 and 5, which contradicts the assumption that p/q was written in lowest terms.