A two digit number is made of two consccutive digits such that the sum of their squares is 4 less than the number. Find the two digit number.

Answer

To solve the problem step by step, we will first define the two-digit number and then set up equations based on the conditions given in the problem.

### Step 1: Define the two-digit number
Let the two-digit number be represented as \( 10x + y \), where \( x \) is the tens digit and \( y \) is the units digit. Since the digits are consecutive, we have two cases:
1. \( x = y + 1 \) (e.g., 12, 23, 34, etc.)
2. \( x = y - 1 \) (e.g., 21, 32, 43, etc.)

### Step 2: Set up the equation based on the problem statement
According to the problem, the sum of the squares of the digits is 4 less than the number itself. Therefore, we can write the equation as:
\[ x^2 + y^2 = (10x + y) - 4 \]

### Step 3: Case 1 - \( x = y + 1 \)
Substituting \( x = y + 1 \) into the equation:
\[ (y + 1)^2 + y^2 = 10(y + 1) + y - 4 \]
Expanding both sides:
\[ (y^2 + 2y + 1) + y^2 = 10y + 10 + y - 4 \]
This simplifies to:
\[ 2y^2 + 2y + 1 = 11y + 6 \]
Rearranging gives:
\[ 2y^2 - 9y - 5 = 0 \]

### Step 4: Solve the quadratic equation
Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 2, b = -9, c = -5 \):
\[ y = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 2 \cdot (-5)}}{2 \cdot 2} \]
Calculating the discriminant:
\[ y = \frac{9 \pm \sqrt{81 + 40}}{4} = \frac{9 \pm \sqrt{121}}{4} = \frac{9 \pm 11}{4} \]
This gives us two possible values for \( y \):
1. \( y = \frac{20}{4} = 5 \)
2. \( y = \frac{-2}{4} = -0.5 \) (not valid since \( y \) must be a digit)

Thus, \( y = 5 \). Now substituting back to find \( x \):
\[ x = y + 1 = 5 + 1 = 6 \]
So, the number is \( 10x + y = 10(6) + 5 = 65 \).

### Step 5: Case 2 - \( x = y - 1 \)
Now we check the second case:
\[ (y - 1)^2 + y^2 = 10(y - 1) + y - 4 \]
Expanding both sides:
\[ (y^2 - 2y + 1) + y^2 = 10y - 10 + y - 4 \]
This simplifies to:
\[ 2y^2 - 2y + 1 = 11y - 14 \]
Rearranging gives:
\[ 2y^2 - 13y + 15 = 0 \]

### Step 6: Solve the second quadratic equation
Using the quadratic formula again:
Here, \( a = 2, b = -13, c = 15 \):
\[ y = \frac{13 \pm \sqrt{(-13)^2 - 4 \cdot 2 \cdot 15}}{2 \cdot 2} \]
Calculating the discriminant:
\[ y = \frac{13 \pm \sqrt{169 - 120}}{4} = \frac{13 \pm \sqrt{49}}{4} = \frac{13 \pm 7}{4} \]
This gives us two possible values for \( y \):
1. \( y = \frac{20}{4} = 5 \)
2. \( y = \frac{6}{4} = 1.5 \) (not valid since \( y \) must be a digit)

Thus, \( y = 5 \) again. Now substituting back to find \( x \):
\[ x = y - 1 = 5 - 1 = 4 \]
So, the number is \( 10x + y = 10(4) + 5 = 45 \).

### Conclusion
The two-digit numbers that satisfy the conditions are **65** and **45**.