Four different integers form an increasing A.P such that one of the them is the same of the square of the remaining numbers . Find the largest number .

Answer

To solve the problem, we need to find four different integers that form an increasing arithmetic progression (A.P.) such that one of the integers is equal to the sum of the squares of the other three integers. Let's denote the four integers in the A.P. as follows:

Let the first term be \( a - d \), the second term be \( a \), the third term be \( a + d \), and the fourth term be \( a + 2d \).

### Step 1: Set up the equation
According to the problem, one of the integers is equal to the sum of the squares of the other three. We can assume that the largest integer \( a + 2d \) is equal to the sum of the squares of the other three integers:

\[ a + 2d = (a - d)^2 + a^2 + (a + d)^2 \]

### Step 2: Expand the squares
Now, we will expand the right-hand side:

\[ (a - d)^2 = a^2 - 2ad + d^2 \]
\[ a^2 = a^2 \]
\[ (a + d)^2 = a^2 + 2ad + d^2 \]

Adding these together:

\[ (a - d)^2 + a^2 + (a + d)^2 = (a^2 - 2ad + d^2) + a^2 + (a^2 + 2ad + d^2) \]
\[ = 3a^2 + 2d^2 \]

### Step 3: Set up the equation
Now substituting back into our equation:

\[ a + 2d = 3a^2 + 2d^2 \]

### Step 4: Rearranging the equation
Rearranging gives us:

\[ 3a^2 - a + 2d^2 - 2d = 0 \]

### Step 5: Solve for \( d \)
This is a quadratic equation in terms of \( d \). We can use the quadratic formula:

\[ d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 2 \), \( b = -2 \), and \( c = 3a^2 - a \).

### Step 6: Discriminant must be non-negative
For \( d \) to be a real number, the discriminant must be non-negative:

\[ (-2)^2 - 4(2)(3a^2 - a) \geq 0 \]
\[ 4 - 8(3a^2 - a) \geq 0 \]
\[ 4 - 24a^2 + 8a \geq 0 \]
\[ -24a^2 + 8a + 4 \geq 0 \]

### Step 7: Factor the quadratic
Dividing the entire inequality by -4 (which flips the inequality):

\[ 6a^2 - 2a - 1 \leq 0 \]

### Step 8: Find roots
Finding the roots of the quadratic \( 6a^2 - 2a - 1 = 0 \):

Using the quadratic formula:

\[ a = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(6)(-1)}}{2(6)} = \frac{2 \pm \sqrt{4 + 24}}{12} = \frac{2 \pm \sqrt{28}}{12} = \frac{2 \pm 2\sqrt{7}}{12} = \frac{1 \pm \sqrt{7}}{6} \]

### Step 9: Determine integer values
The roots give us intervals for \( a \). Since \( a \) must be an integer, we can test integer values within the range defined by the roots.

### Step 10: Testing values
Testing \( a = 0 \):

\[ d = 1 \quad \text{(as calculated earlier)} \]
The integers are \( -1, 0, 1, 2 \).

### Conclusion
The largest integer in this A.P. is:

\[ \boxed{2} \]