Question

Determine all prime numbers a, b and c for which the expression a ^ 2 + b ^ 2 + c ^ 2 - 1 is a perfect square .

I need the answer step by step

Answer 1

The family of all prime numbers such that
`a^(2) + b^(2) + c^(2) -1` is a perfect square is represented by the following solution:

`a` is an arbitrary prime number. (1)

`b = √(1 + 2\cdot a \cdot c)` (2)

`c` is another arbitrary prime number. (3)

Explanation:

From Algebra we know that a second order polynomial is a perfect square if and only if
`(x+y)^(2) = x^(2) + 2\cdot x\cdot y + y^(2)`. From statement, we must fulfill the following identity:

`a^(2) + b^(2) + c^(2) - 1 = x^(2) + 2\cdot x\cdot y + y^(2)`

By Associative and Commutative properties, we can reorganize the expression as follows:

`a^(2) + (b^(2)-1) + c^(2) = x^(2) + 2\cdot x \cdot y + y^(2)` (1)

Then, we have the following system of equations:

`x = a` (2)

`(b^(2)-1) = 2\cdot x\cdot y` (3)

`y = c` (4)

By (2) and (4) in (3), we have the following expression:

`(b^(2) - 1) = 2\cdot a \cdot c`

`b^(2) = 1 + 2\cdot a \cdot c`

`b = √(1 + 2\cdot a\cdot c)`

From Number Theory, we remember that a number is prime if and only if is divisible both by 1 and by itself. Then,
`a, b, c > 1`. If
`a`,
`b` and
`c` are prime numbers, then
`2\cdot a\cdot c` must be an even composite number, which means that
`a` and
`c` can be either both odd numbers or a even number and a odd number. In the family of prime numbers, the only even number is 2.

In addition,
`b` must be a natural number, which means that:

`1 + 2\cdot a\cdot c \ge 4`

`2\cdot a \cdot c \ge 3`

`a\cdot c \ge (3)/(2)`

But the lowest possible product made by two prime numbers is
`2^(2) = 4`. Hence,
`a\cdot c \ge 4`.

The family of all prime numbers such that
`a^(2) + b^(2) + c^(2) -1` is a perfect square is represented by the following solution:

`a` is an arbitrary prime number. (1)

`b = √(1 + 2\cdot a \cdot c)` (2)

`c` is another arbitrary prime number. (3)

Example:
`a = 2`,
`c = 2`

`b = √(1 + 2\cdot (2)\cdot (2))`

`b = 3`