Basic Math Problem of the Week 11/14/2017

[PF PotW Robot]

Here is this week's basic math problem. We have several members who will check solutions, but we also welcome the community in general to step in. We also encourage finding different methods to the solution. If one has been found, see if there is another way. Using spoiler tags is optional. Occasionally there will be prizes for extraordinary or clever methods.

Positive integers $a,\,b, \,c, \,d$, and $e$ satisfy the equations

$$(a + 1)(3bc + 1) = d + 3e + 1\\
(b + 1)(3ca + 1) = 3d + e + 13\\
(c + 1)(3ab + 1) = 4(26 − d − e) − 1$$

Evaluate $d^2 + e^2$.

(PotW thanks to our friends at http://www.mathhelpboards.com/)

 

[fresh_42]

Hint: Observe that $3 | (a+b+c)$.

 

[ddddd28]

The right hand of the third equation says that d+e is less than 26.
Let d+e =25 , we get 3 . 3 is not possible because the left hand of the third equation can be minimum 8.( a=b=c=1)
d+e=24 is not possible for the same reasons.
Substituting d+e= 23, 21,20,18,15,14,11,9,8,6,5 gives a prime number- contradiction.
If d+e is 22, we get 15, which can be written as 3*5 or 5*3. However 3ab+1 is only one of these: 4,7,10,13,16... It means that 15 cannot be expressed by the left hand. So d+e is not 22.
For the same reasons, d+e is not equal to 19, 13, 12,4.
If d+e is 2 then d=1 e=1 . Plugging it in the first equation gives 5 - prime number.
d+e=1 is not possible because d and e are natural.
We are left with 17, 16,10,7 and 3.
I will explain how I delete them only with one number because it's too long to write it down.
d+e is 10. It gives 63 in the right hand of the third equation. 63 can be expressed as:
3,21
7,9
9,7
21,3
Assume the first number in order is 3ab+1 and the second is c+1.
As mentioned earlier, 3ab+1 is only one of these : 4,7,10,13...
So only 7,9 remains. 9=c+1 and 7 =3ab+1.
So, the triples that satisfy it are (2,1,8) and (1,2,8). ( The order is a,b,c)
Recall that fresh_42 said that a+b+c= 3k, k is a natural number. This fact can be achieved by adding all the equations together.
At last, we can delete these triples because of that fact. Concluding that d+e is not 10.
Doing so to the remaining numbers ( which didn't take more than 10 minutes), leaves 16 with the triple (2,2,2) and 3 with the triples (2,1,12) and ( 1,2,12).
Now all that has to be done is solving a system of two equations, that is to say, d+e=m and the first equation ( for instance) for the three triples.
Solving for 3 gives a negative d, so only 16 is left.
Finally, we get d=5 and e=11 and a=b=c=2.
My answer is d^2+e^2=146.