Practice! #4
Positive integers , and , where , satisfy the system of equations above. If there are triplets then enter as your answer.
The answer is 54.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
given,
a 2 + b 2 = c 2 ,
( a b / 2 ) = a + b + c ,
( a b / 2 ) - c = a + b ,
Squaring on both sides,
We get ,
( a 2 b 2 / 4 ) + c 2 − a b c = a 2 + b 2 + 2 a b = c 2 + 2 a b
( a 2 b 2 / 4 ) − a b c = 2 a b . Now divide by a b on both sides.
( a b / 4 ) - c = 2
( a b / 4 ) - ( ( a b / 2 ) − a − b ) = 2
a + b − ( a b / 4 ) = 2.
( a − 4 ) ( b − 4 ) = 8.
So now, a − 4 can be 1 , 2 , 4 , − 1 , − 2 , − 4 , 8 , − 8 .
By putting ′ a ′ values
We get ( a , b ) as ( 5 , 1 2 ) , ( 6 , 8 ) , ( 8 , 6 ) , ( − 3 , − 1 2 ) , ( − 2 , 0 ) , ( 0 , − 2 ) , ( 1 2 , 5 ) , ( − 4 , 3 )
But a < b < c and c is an integer. And all are positive ,
So only solutions we get are ( 5 , 1 2 , 1 3 ) , ( 6 , 8 , 1 0 ) as ( a , b , c ) respectively.
So sum of all possible values is ( a , b , c ) is, 5 + 1 2 + 1 3 + 6 + 8 + 1 0 = 5 4 .