Problem 1497

1. Definition and Initial Setup:
We need to prove that there are infinitely many consecutive quadruples of almost squares. A natural number $n$ is called *almost a square* if it can be represented as a product of two numbers that differ by no more than one percent of the larger of them.

2. Identifying Almost Squares:
Consider the numbers $4m^4 - 1$, $4m^4$, and $4m^4 + 1$. We need to show that these numbers are almost squares for large $m$.

3. **Verification for $4m^4 + 1$**:
$$4m^4 + 1 = (2m^2 - 2m + 1)(2m^2 + 2m + 1)$$
We need to check that the ratio of the factors is close to 1:
$$\frac{2m^2 - 2m + 1}{2m^2 + 2m + 1} > 0.99 \quad \text{for} \quad m \geq 199$$
This inequality holds for sufficiently large $m$, specifically $m \geq 199$.

4. **Verification for $4m^4$**:
$$4m^4 = (2m^2)(2m^2)$$
Clearly, $4m^4$ is a perfect square and thus trivially an almost square.

5. **Verification for $4m^4 - 1$**:
$$4m^4 - 1 = (2m^2 - 1)(2m^2 + 1)$$
We need to check that the ratio of the factors is close to 1:
$$\frac{2m^2 - 1}{2m^2 + 1} > 0.99 \quad \text{for} \quad m \geq 199$$
This inequality also holds for sufficiently large $m$.

6. **Verification for $4m^4 + 2$**:
We need to show that $4m^4 + 2$ is an almost square. Assume:
$$4m^4 + 2 = (2m^2 - 2a)(2m^2 + b)$$
Solving for $m^2$:
$$m^2 = \frac{ab + 1}{b - 2a}$$
Let $b - 2a = 1$:
$$m^2 = 2a^2 + a + 1$$
This is a Pell's equation, which has infinitely many solutions starting with $(0,1)$ and $(1,2)$.

7. Conclusion:
Therefore, $4m^4 + 2 = (4a^2 + 2)(4a^2 + 4a + 3)$ is an almost square for infinitely many values of $a$.

Thus, we have shown that there are infinitely many consecutive quadruples of almost squares.

$\blacksquare$