# The sum of the squares of two consecutive positive numbers is 61. What is the smaller number?

Jun 28, 2016

Smallest no is $5$. Other no is 6.

#### Explanation:

Let the smaller no be $n$. So the other no is $n + 1$

Given: ${n}^{2} + {\left(n + 1\right)}^{2} = 61 \implies$

${n}^{2} + {n}^{2} + 2 n + 1 - 61 = 0$

${n}^{2} + {n}^{2} + 2 n + 1 - 61 = 0$

$2 {n}^{2} + 2 n - 60 = 0$

${n}^{2} + n - 30 = 0$

$\left(n + 6\right) \left(n - 5\right) = 0$ =>  n = -6$\mathmr{and}$n=5#

Discard $- 6$ as n is positive. So, $n = 5$

Jun 28, 2016

The smaller number is $5$.

#### Explanation:

Let the smaller number be $x$ and then next consecutive number is $x + 1$. As the sum of the squares of these two consecutive positive numbers is $61$, we have

${x}^{2} + {\left(x + 1\right)}^{2} = 61$ or

${x}^{2} + {x}^{2} + 2 x + 1 = 61$ or

$2 {x}^{2} + 2 x - 60 = 0$ or (diving each side by $2$)

${x}^{2} + x - 30 = 0$ or

${x}^{2} + 6 x - 5 x - 30 = 0$ or

$x \left(x + 6\right) - 5 \left(x + 6\right) = 0$ or

$\left(x + 6\right) \left(x - 5\right) = 0$

Hence $x = - 6$ or $x = 5$.

But as we need positive numbers only, $x = 5$.