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Given the system of equations:
$a^2 = 9b+10c+11d$
$b^2 = 9c+10d+11a$
$c^2 = 9d+10a+11b$
$d^2 = 9a+10b+11c$

How do I prove that the only positive real solution is {30, 30, 30, 30}?

So far, I've tried adding all of the equations up and then manipulating them but that didn't end up helping me disprove the existence of other positive real solutions. I've also tried subtracting the equations from each other and even using the quadratic formula to try to get a relation between the variables independently.

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    "I've tried adding all of the equations up" --- and just what did you get when you did that? – John Hughes Oct 01 '20 at 00:11
  • Well, you get $a^2 + b^2 + c^2 +d^2 = 30(a + b +c +d)$ which makes it pretty clear that {30, 30, 30, 30} is a solution. However, I'm getting stuck at disproving that there aren't any other solutions. –  Oct 01 '20 at 00:14
  • Why just non-negative integers must be checked @Will Jagy ? – Tortar Oct 01 '20 at 00:47

2 Answers2

2

Approach 1: Adding up the equations, $30 (a + b + c + d) = a^2 + b^2 + c ^2 + d^2 \geq \frac{ (a+b+c+d) ^2 } { 4} $.
Since $ a+b + c + d > 0$, we can divide by it and conclude that hence $ 120 \geq a+b+c+d$.

WLOG, let $ a= \min (a, b, c, d)$.
Then $ a^2 = 9b + 10c + 11d \geq 9a + 10 a + 11a = 30a$, so $ a \geq 30$.

The only possible tuple which satisfies both conditions is $(30,30,30,30)$, which we can easily verify is a solution.

Approach 2: (Fill in the gaps yourself) Similar to the 2nd step from above, show that

$$ 30 \leq \min (a, b, c, d) \leq \max (a, b, c, d ) \leq 30. $$

Note:

  1. I believe (no proof yet) that $ (0,0,0,0)$ is the only other real solution.
  2. Wolfram says that there are other complex solutions.
Calvin Lin
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2

Let $a=\min\{a,b,c,d\}$.

Thus, $$a^2=9b+10c+11d\geq30a,$$ which gives $a\geq30.$

In another hand, by C-S $$\sum_{cyc}a^2=30(a+b+c+d)\leq30\sqrt{4(a^2+b^2+c^2+d^2)},$$ which gives $$a^2+b^2+c^2+d^2\leq3600,$$ which says $$a=b=c=d=30.$$

Michael Rozenberg
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