Let a and b be positive integers. Find a+b if a2+a=b4+b3+b2+b

We are given the equation:

\[
a^2 + a = b^4 + b^3 + b^2 + b
\]

where \(a\) and \(b\) are positive integers. Our goal is to find \(a + b\).

### Step 1: Factor the right-hand side

We start by factoring the right-hand side of the equation:

\[
b^4 + b^3 + b^2 + b = b(b^3 + b^2 + b + 1)
\]

This simplifies the equation to:

\[
a^2 + a = b(b^3 + b^2 + b + 1)
\]

### Step 2: Try small integer values of \(b\)

We will now substitute small values of \(b\) and see if we can find a corresponding \(a\) that satisfies the equation.

#### Case 1: \(b = 1\)

Substituting \(b = 1\) into the equation:

\[
b^4 + b^3 + b^2 + b = 1^4 + 1^3 + 1^2 + 1 = 1 + 1 + 1 + 1 = 4
\]

So, the equation becomes:

\[
a^2 + a = 4
\]

We solve the quadratic equation:

\[
a^2 + a - 4 = 0
\]

Using the quadratic formula:

\[
a = \frac{-1 \pm \sqrt{1^2 - 4(1)(-4)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 16}}{2} = \frac{-1 \pm \sqrt{17}}{2}
\]

Since \(\sqrt{17}\) is not an integer, there is no solution for \(a\) when \(b = 1\).

#### Case 2: \(b = 2\)

Substituting \(b = 2\) into the equation:

\[
b^4 + b^3 + b^2 + b = 2^4 + 2^3 + 2^2 + 2 = 16 + 8 + 4 + 2 = 30
\]

So, the equation becomes:

\[
a^2 + a = 30
\]

We solve the quadratic equation:

\[
a^2 + a - 30 = 0
\]

Using the quadratic formula:

\[
a = \frac{-1 \pm \sqrt{1^2 - 4(1)(-30)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 120}}{2} = \frac{-1 \pm \sqrt{121}}{2} = \frac{-1 \pm 11}{2}
\]

This gives two possible solutions for \(a\):

\[
a = \frac{-1 + 11}{2} = 5 \quad \text{or} \quad a = \frac{-1 - 11}{2} = -6
\]

Since \(a\) must be a positive integer, we take \(a = 5\).

### Step 3: Find \(a + b\)

Now that we have \(a = 5\) and \(b = 2\), we can find \(a + b\):

\[
a + b = 5 + 2 = 7
\]

Thus, the value of \(a + b\) is \( \boxed{7} \).

---

Would you like more details or have any questions?

Here are 5 related questions:
1. What happens if \(b = 3\) is tried?
2. How do you solve quadratic equations using factoring?
3. Can this equation be generalized for higher powers of \(b\)?
4. How would the solution change if \(a\) and \(b\) were not integers?
5. What if we tried different methods of factoring the original equation?

**Tip**: When solving quadratic equations, always check if factoring is easier before using the quadratic formula.

Let a and b be positive integers. Find a+b if a^2+a=b^4+b^3+b^2+b.

Find a+b if a^2 + a = b^4 + b^3 + b^2 + b, where a and b are positive integers. This problem involves algebraic factoring and solving quadratic equations. Suitable for students in Grades 9-11.

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