# Explain using algebra,why the difference of diagonal pair in a square of calendar is always 14.

Let’s use algebra to see this.

Taking the first number in the square as x, the others can be filled as below

the squares of diagonal are

x^{2} and (x+8)^{2}

(x+7)^{2} and (x+1)^{2}

Add the squares of the diagonal pair

x^{2} + (x+8)^{2}

= x^{2} + (x^{2}+8^{2}+2 [Using identity = + + 2]

^{=} 2x^{2} + 64+16

Add the squares of the other diagonal pair

= (x+1)^{2}+(x+7)^{2}

^{=} (x^{2} +1 +2 (x^{2}+7^{2}+2

= (x^{2} +1 +2 (x^{2} +49 +14)

= 2x^{2} +16+ 50

find the difference of these sums:

= (2x^{2} + 64+16- (2x^{2} +16+ 50)

= 2x^{2} + 64+16- 2x^{2} -16- 50

= 64-50

= 14

Hence the difference is 14, we can take any number as x; which means this holds in any part of the calendar.