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, we need to find a two-digit number made of two consecutive digits such that the sum of their squares is 4 less than the number itself. Let's break down the solution step by step.

### Step 1: Define the digits
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. \(x = y + 1\) (the tens digit is greater)
2. \(x = y - 1\) (the units digit is greater)

### Step 2: Set up the equation
According to the problem, the sum of the squares of the digits is equal to the number minus 4:
\[ x^2 + y^2 = (10x + y) - 4 \]
This simplifies to:
\[ x^2 + y^2 = 10x + y - 4 \]

### 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, \(a = 2\), \(b = -9\), and \(c = -5\):
\[ y = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 2 \cdot (-5)}}{2 \cdot 2} \]
Calculating the discriminant:
\[ y = \frac{9 \pm \sqrt{81 + 40}}{4} = \frac{9 \pm \sqrt{121}}{4} = \frac{9 \pm 11}{4} \]
Calculating the two possible values for \(y\):
1. \(y = \frac{20}{4} = 5\)
2. \(y = \frac{-2}{4} = -0.5\) (not valid)

Thus, \(y = 5\) and substituting back gives \(x = 6\).

### Step 5: Case 2: \(x = y - 1\)
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:
\[ 2y^2 - 2y + 1 = 11y - 14 \]
Rearranging gives:
\[ 2y^2 - 13y + 15 = 0 \]

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

Thus, \(y = 5\) again gives \(x = 4\).

### Step 7: Conclusion
The two-digit numbers we found are:
1. From Case 1: \(65\)
2. From Case 2: \(45\)

Thus, the two-digit numbers that satisfy the condition are **65** and **45**.