Question

find the two consecutive positive integers,sum of whose squares is 613

Answer 1

take 1st number = x

2nd number = x+1

x^2 + ( x + 1 )^2 = 613

x^2 + x^2 + 2x + 1 = 613

2x^2 + 2x = 613-1

2 ( x^2 + x ) = 612

x^2 + x = 612 ÷ 2

x^2 + x = 306

x^2 + x -306 = 0 [ make it as a quadratic equation ]

-b + or - √ b^2 - 4ac /2a  [ a = 1 , b = 1 , c = -306 ]

-1 + or - √ 1^2 + (4×1×306) /2

-1 + or - 35/2

-1+35/2 = 34/2

first number = 17

so the second number = 18

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Assumption

First positive integers be t.

Hence,

Next integer will be t + 1

${\boxed{\sf\:{Situation}}}$

t² + t² + 2 × t × 1 + 1² = 613

t² + t² + 2t + 1 - 613 = 0

2t² + 2t - 612 = 0

2(t² + t - 306) = 0

t² + t - 306 = 0

${\boxed{\sf\:{Solving\;the\;Quadratic\;Equation}}}$

t² + 18t - 17t - 143 = 0

t(t + 18 ) - 17 (t + 18 ) = 0

(t - 17)(t + 18) = 0

t - 17 = 0

t = 17

t + 18 = 0

t = -18

${\boxed{\sf\:{t\;is\;an\;odd\;positive\;integer}}}$

Hence,

t ≠ - 18

t = 17

Therefore,

${\boxed{\sf\:{Two\;consecutive\;positive\;integers}}}$

${\boxed{\sf\:{17\;and\;18}}}$