Question 189204
(1)
Call the 3 integers {{{a}}}, {{{a + 1}}}, and {{{a + 2}}}
{{{a^2 + (a + 1)^2 = (a + 2)^2}}}
{{{a^2 + a^2 + 2a + 1 = a^2 + 4a + 4}}}
Subtract {{{a^2}}} from both sides
{{{a^2 + 2a + 1 = 4a + 4}}}
Subtract {{{4a}}} from both sides
{{{a^2 - 2a + 1 = 4}}}
Subtract {{{4}}} from both sides
{{{a^2 - 2a - 3 = 0}}}
{{{(a - 3)(a + 1) = 0}}}
{{{a = 3}}}
{{{a = -1}}} Can't use the negative root
{{{a = 3}}}
{{{a + 1 = 4}}}
{{{a + 2 = 5}}}
The consecutive numbers are 3,4, and 5
(2)
If the integers are odd, they are every other number:
{{{a}}}, {{{a + 2}}}, {{{a + 4}}}
{{{a + a + 2 + a + 4 = a*(a + 2) - 36}}}
{{{3a + 6 = a^2 + 2a - 36}}}
Subtract {{{3a}}} from both sides
{{{6 = a^2 - a - 36}}}
Subtract {{{6}}} from both sides
{{{a^2 - a - 42 = 0}}}
{{{(a - 7)(a + 6) = 0}}}
{{{a = 7}}}
{{{a = -6}}} Can't use negative root
{{{a = 7}}}
{{{a + 2 = 9}}}
{{{a + 4 = 11}}}
the consecutive odd integers are 7, 9, and 11
check:
{{{a + a + 2 + a + 4 = a*(a + 2) - 36}}}
{{{7 + 9 + 11 = 7*9 - 36}}}
{{{27 = 63 - 36}}}
{{{27 = 27}}}
OK