Let 3<em>n</em> + 1 denote the "number" in question. The claim is that
(3<em>n</em> + 1)² = 3<em>m</em> + 1
for some integer <em>m</em>.
Now,
(3<em>n</em> + 1)² = (3<em>n</em>)² + 2 (3<em>n</em>) + 1²
… = 9<em>n</em>² + 6<em>n</em> + 1
… = 3<em>n</em> (3<em>n</em> + 2) + 1
… = 3<em>m</em> + 1
where we take <em>m</em> = <em>n</em> (3<em>n</em> + 2).
Answer:
Step-by-step explanation:
No because √2 and 3i come in pairs. The four roots are ±√2, 3i and -3i. The 5 guarantees that there are at least 5 roots to this equation. This is a good question to know the answer to. It looks like something that could be put on a test.
B+14/2=b+14/2
1
they r equal
The answer to your question would be 'A'
Answer:
Can you get a picture please or more detail...
