How to Find All Natural Numbers n Such That (n^2 ab) and (n^3 a^2b^2)

To find all natural numbers n such that n2 ab and n3 a2b2 where a and b are integers, we can start by manipulating these equations.

Expressing b in Terms of a

From the first equation, we can express b in terms of n and a:

b n2 - a.

Substituting b into the Second Equation

Now, substitute this expression for b into the second equation:

n3 a2(n2 - a)^2.

Expanding the right-hand side:

n3 a2(n4 - 2n2a a2)

Rearranging gives us:

0 n4 - n3 - 2n2a - a2.

Isolating a

This can be rearranged to isolate a:

2a2 - 2n2a - n4 n3 0.

This is a quadratic equation in a. Applying the quadratic formula:

a frac{-B pm sqrt{B^2 - 4AC}}{2A}

Here, A 2, B -2n2, and C n4 - n3. Substituting these values in:

a frac{2n2 pm sqrt{4n4 - 8n4 - (n4 - n3)}}{4} frac{2n2 pm sqrt{4n4 - 8n4 2n3 - n4}}{4}

The expression simplifies to:

a frac{n2 pm sqrt{2n3 - n4}}{2}.

Conditions for a to Be an Integer

For a to be an integer, 2n^3 - n^4 must be a perfect square. Let's denote:

k2 n4 - 2n3.

Rearranging gives:

n4 - 2n3 - k2 0.

Using the discriminant:

D 4n6 - 4k2 4n6 - 4k2.

This must also be a perfect square.

Testing Small Values of n

We can test small natural numbers to find valid pairs a, b.

For n 1: 12 ab implies ab 1 13 a2b2 implies a2b2 1.

The pairs 0, 1, 1, 0 work so n 1 is a solution.

For n 2: 22 ab implies ab 4 23 a2b2 implies a2b2 8.

Trying a 0, 4 gives no valid pairs.

For n 3: 32 ab implies ab 9 33 a2b2 implies a2b2 27.

Testing valid pairs yields no results.

Continuing this process, we can conclude that the only natural number n that satisfies both equations is:

boxed{1}.