perfect square

Reasoning does the trick .

From your provided equation , try to get to the following result -

( b - c ) ( a - c ) = c ²

Let " P " be a prime which divides " c " .

Then obviously , " P ² " will divide " c ² " , i . e , " P ² " will divide " ( a - c ) ( b - c ) " . But , if

" P ² " divides both of them , then there will arise a contradiction --

since , a - cP = aP - cP <----- this must be an integer , so " P " must divide " a " , as our early

assumption was that " P " divides " c " . But " P " cannot divide both of them , as they don ' t

have any common factor . So " P ² " will either divide " ( a - c ) " or " ( b - c ) " .

Hence , either " ( a - c ) " or " ( b - c ) " is a perfect square . But if one is a perfect square ,

then the other must also be a perfect square , since their product is a square = c ² .

Now , choose , a - c = k ² , and b - c = c ²k ² , where obviously " k " divides " c " .

So , ( k + ck ) ² = k ² + 2 c + c ²k ² = a - c + b - c + 2c = a + b

As , " k " divides " c " , hence " ( k + ck ) ² " must also be an integer , rather a square of an integer .

So , " a + b " is an integer which is a perfect square .