Primes Question - AMTI Inter Level Finals

If $x = 2$ then $x(x + y)$ is even. For $z + 120$ to then also be even for $z$ prime will require that $z = 2.$ This in turn requires that

$2(2 + y) = 2 + 120 \Longrightarrow 2 + y = 61 \Longrightarrow y = 59.$

Since $59$ is prime we have found a solution $(x,y,z) = (2,59,2).$

Now if prime $x \ne 2$ then if $y$ is an odd prime we would have $x + y$ being even, implying that $x(x + y)$ would also be even. But $z + 120$ with $z$ prime can only be even if $z = 2,$ and since $122 = 2*61$ and $x \lt x + y$ the only way we could have $x(x + y) = 122$ is to have $x = 2.$ Thus if $x \ne 2$ we will require that $y = 2$ if we are to find any additional prime solutions for $z.$

So we now need to look at the equation $x(x + 2) = z + 120 \Longrightarrow x^{2} + 2x - (z + 120) = 0.$ This has solutions

$x = \dfrac{-2 \pm \sqrt{4 + 4(z + 120)}}{2} = -1 \pm \sqrt{z + 121}.$

As we require that $x \gt 0$ we choose the positive root, and since we also require that $x$ is an integer we must have that $z + 121 = n^{2}$ for integer $n.$ This can be written as

$z = n^{2} - 121 = n^{2} - 11^{2} = (n - 11)(n + 11).$

For $z$ to be prime we must then have that either $n - 11 = 1$ or $n + 11 = 1,$ resulting in respective values for $n$ of $12$ or $-12.$ Both these values give us that $z = 12^{2} - 121 = 23,$ and thus $x = -1 + 12 = 11,$ both of which are prime. Thus we have found a second solution $(x,y,z) = (11,2,23).$

As this completes the casework, we can conclude that the only solution triples are $(2,59,2)$ and $(11,2,23),$ and so the desired answer is $(2 + 59 + 2) + (11 + 2 + 23) = \boxed{99}.$

My solution is very similar to Brian Charlesworth's. The main difference is in the handling of the y=2 case.

If y=2 , then

x(x+2)=z+120

x^2 + 2x - 120 = z

(x - 10)(x+12)=z

As z is prime, it can only have two factors: 1 and itself. Therefore, either x-10=1 or x+12=1. As z>0, the only solution of the y=2 case is x=11 and z=23.

I solved it as follows:

$(i)$ Since all prime no.'s must be odd except 2, our first step should be to observe the given expression. If $y\neq2$ , LHS will always be even & thus for RHS to be even z must be even, i.e. $z=2$ .
So let's solve for this case. We have,

$x(x+y)=122=2*61$

comparing the two sides, we get only one solution, i.e. x=2 and y=59.

So our first set of values for (x,y,z) becomes $(2,59,2).$

$(ii)$ Now let's look at the case when y=2. We have,

$x(x+2)=z+120$

$\Rightarrow z=x^{2} + 2x - 120$

$\Rightarrow z=(x+1)^{2} - 11^{2}$

$\Rightarrow z=(x+12)(x-10)$

We have written z, which is a prime no., as a product of two no.'s which is not possible unless one of them equals 1.

$\Rightarrow (x-10)=1$

and hence x=11. Putting values of x & y we get, z=23.

So our second set of values for (x,y,z) are $(11,2,23).$

$(iii) Now\ required\ sum= (2+11)+(59+2)+(2+23)=99 \ \ \ \ Ans.$

Scientific approach.

2 (2 + 59) = 2 + 120 = 122

11 (11 + 2) = 23 + 120 = 143

Checked with first 1754 primes for up to 15000 of prime 14983, despite mathematical report.

2 + 59 + 2 + 11 + 2 + 23 = 99