Q13 of 168 Page 7

The sum of the squares of two consecutive natural numbers is 421. Find the numbers.

Let the first number be ‘X’, so the other number will be ’(X+1)’.


∵ X2 + (X + 1)2 = 421


X2 + X2 + 1 + 2X = 421


On simplifying further,


2X2+ 2X – 420 = 0


X2 + X – 210 = 0


On applying Sreedhracharya formula







∴ X = –15 or X = 14.


∴ X = 14 (Only natural number, as given in the question)


∴ Then other numbers will be 14 & 15.


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