**Problem 1 :**

Find two consecutive positive even integers whose squares have the sum 340.

**Solution :**

Let x and (x + 2) be two positive even integers.

Sum of their squares = 340

x^{2} + (x + 2)^{2} = 340 ---(1)

Expanding (x + 2)^{2 }using the the algebraic identity

(a + b)^{2} = a^{2} + 2ab + b^{2}

(x + 2)^{2} = x^{2} + 2 (x) (2) + 2^{2}

(x + 2)^{2 }= x^{2} + 4x + 4

By applying the expansion in (1), we get

x^{2} + x² + 4x + 4 = 340

2x² + 4x - 336 = 0

By dividing the entire equation y 2, we get

x^{2} + 2x - 168 = 0

By factoring, we get

(x - 12) (x + 14) = 0

x - 12 = 0 x = 12 |
x + 14 = 0 x = -14 |

Since the required number is even positive integers, we choose positive integer.

If x = 12, then x + 2 ==> 14

Therefore two positive even integers are 12 and 14.

**Verification :**

Sum of their squares is 340

= 12^{2} + 14^{2}

= 144 + 196

= 340

**Problem 2 :**

The sum of two squares of three consecutive natural numbers is 194. Determine the numbers.

**Solution :**

Let x , x + 1 and x + 2 be three consecutive natural numbers

Sum of their squares = 194

x^{2} + (x + 1)^{2} + (x + 2)^{2} = 194

(x + 1)^{2 }= x^{2} + 2x + 1

(x + 2)^{2 }= x^{2} + 4x + 4

x^{2} + x^{2} + 2x + 1 + x^{2} + 4x + 4 = 194

3x^{2} + 6x + 5 = 194

3x^{2} + 6x + 5 - 194 = 0

3x^{2} + 6x - 189 = 0

Dividing the entire equation by 3, we get

x^{2} + 2x - 63 = 0

(x + 9) (x - 7) = 0

x + 9 = 0 x = -9 |
x - 7 = 0 x = 7 |

If x = 7

then x + 1 ==> 7 + 1 ==> 8

and x + 2 ==> 7 + 2 ==> 9

Therefore the required numbers are 7, 8, 9

**Verification :**

The sum of two squares of three consecutive natural numbers is 194

= 7^{2} + 8^{2} + 9^{2}

= 49 + 64 + 81

= 194

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