krusta80 wrote:Interesting side question I posed to myself while looking at part 2 of this question is (and I warn anyone trying to stay on topic to AVOID the rest of this post):
How many square integers, x, exist such that 3x+4 is also a perfect square?
This is a great Q.
Since x is a perfect square, we can write x as k², where k is some integer. In these terms, 3x + 4 is 3k² + 4. This is also a perfect square; call it m². This gives us the equation
3k² + 4 = m²
3k² = m² - 4
3k² - m² + 4 = 0
Let's treat this as a quadratic in k, so that the a-coefficient = 3, the b-coefficient is 0, and the c-coefficient is (4 - m²).
This gives us k = ±√(0² - 4*3*(4-m²)) / 6, or ±√(12m² - 48)/6, or ±2*√(3m² - 12)/6, or ±√(3m² - 12)/3.
So provided that √(3m² - 12)/3 is an integer, we'll have a solution. To find some of these solutions, let's begin with an easy one: m = 2, which gives √(3m² - 12) = 0, an integer multiple of 3. Another easy one is m = 4, which gives √(3m² - 12) = 6, another integer multiple of 3.
In fact -- and I can't find a way to represent this next step in arithmetic appropriate to the GMAT -- we can go further, proceed inductively, and find more of solutions.
Let's start with our two basic solutions, m = 2 and m = 4. Notice that 4*4 - 2 = 14 is another solution: √(3*14² - 12) = 24, again an integer multiple of 3. Now we have m = 14 as a solution. Observe that 4*14 - 4 = 52 is another solution, as is 4*52 - 14 = 194, as is 4*194 - 52 = 724, etc. The inductive step would require us to prove that if r and s are "consecutive" solutions, then (4s - r) is also a solution, but I don't know a nice way of doing this in GMAT-friendly terms.