Find all integer solutions of the equation a^2 + b^2 + c^2 = 2abc?

Question

Curt, why do you say that p is odd?

Answer 1

Assume that a, b, and c are all positive and have no common factors greater than 1. Therefore, at least one of a, b, or c is odd. Furthermore, if two of a, b, c are even, then so is the remaining one since (assuming a, b are both even) if c² = 2abc - a² - b² the terms on the right are all even. Thus, at least two of a, b, c are odd. Clearly, all three can't be odd since then the left hand side would be odd. Therefore, exactly two of a, b, c are odd and one is even.

Therefore, assume that a = 2x-1 for some positive integer x, and that b = 2y-1 for some positive integer y, and that c=2z for some positive integer z. Then we are looking for solutions of

(2x-1)² + (2y-1)² + 4z² = 2abc
Expanding yields:
4x² - 4x +1 + 4y² - 4y + 1 + 4z² = 4abz
Simplifying yields:
4x² + 4y² + 4z² - 4x - 4y - 4abz = 2

The left side is clearly a multiple of 4, while the right side is not, a contradiction. Therefore, there are no solutions.

*****
Regarding Zanti's comments, he is correct that there is a mistake above. Rather than edit the solution in place, I'm going to point out what needs to be changed so that the difference is readily apparent.


Firstly, I didn't cover non positive solutions: If any of a,b, or c is 0, then 2abc is 0 so all three of a, b, c must be 0. Also, if one of the triple is negative, then exactly one other component is negative (so that the right side is positive). Thus, forming a triple with the absolute values of each of the triples' components produces an all positive triple. Similarly, if there is a solution of all positive a, b, c, then any two of them being negative would also be a solution. Thus, to find all solutions, we can look at all positive only solutions.

Secondly, this is not a proof by induction. It's a proof by contradiction, without an inductive component (Induction is where you show the truth of the statement for some initial value(s), and then show that the assumption of truth for an arbitrary value implies the truth for subsequent values).


Most importantly, My argument above actually proves a stronger statement, when corrected. Specifically, I show that there are NO non 0 triples (a, b, c) such that
a² + b² + c² = (2^s)abc for any integers s>=1

Now if (a,b,c)=2^t(p,q,r) were to satisfy the above where t is a non negative integer then, as Zanti has pointed out, we arrive to p² + q² + r² = (2^t)(2^s)pqr = (2^(s+t))pqr

Thus, I can assume a triple (a,b,c) that has no common factor of 2, so that at least one of the triple must be odd. And clearly an even number of the triple components must be odd since the left side is even. Hence, exactly two elements of (a,b,c) are odd.

And now the proof continues from the second paragraph of my original text above, only one must substitute
(2^s)abc for s>=1 for 2abc and
(2^(s+1))abz for s>=1 for 4abz

By the way, that s>=1 is necessary. If s=0 then the triple (3,3,3) or (3,3,6) satisfies
a² + b² + c² = abc

Answer 2

If my analysis is right, the only integral solution is the trivial a = b = c = 0. To prove this:

If exactly one of the numbers is odd, then a^2 + b^2 + c^2 is odd, while 2abc is even. Hence, no solution.

If all three of the numbers are odd, then a^2 + b^2 + c^2 is odd, while 2abc is even. Hence, no solution.

If exactly two of the numbers are odd, then a^2 + b^2 + c^2 is divisible by 2 but NOT divisible by 4. To prove this, note that if you square an odd number and then divide it by 4, the remainder is 1. Therefore if you sum the squares of two odd numbers and one even number, then divide this sum by 4, the remainder must be 2. However, 2abc must be divisible by 4. So, again there can be no solution.

That means a, b, and c must all be even numbers. Let's let a = 2p, b = 2q, and c = 2r. Then, your equation can be rewritten as 4(p^2 + q^2 + r^2) = 16pqr, or p^2 + q^2 + r^2 = 4pqr. This equation is not much different from your original equation, and we can use the same arguments to prove p, q, and r must also be even. So, we can make another substitution, e.g. s = 2p, t = 2q, and u = 2r.

So, you can see what is happening: each new substitution yields a new equation where an integral solution has to be an even number, but the numbers are half as large after each new substitution. Assuming at least one number is positive, eventually you have to get to a point where a number is not even. (For example, if all the numbers are powers of 2, eventually one of the numbers is reduced to 1.) That leaves a = b = c = 0 as the only possibility.

* * * * *

Hmm... I'm looking at quadrillerator's post above me. If you disregard the all zero solution, we're on the same wave length; however, I don't think he can validly assume a, b, and c have no common factors. For example, assume a = kp, b = kq, and c = kr for some integer k > 1. Then, to prove this inductively, we would proceed:

Assume a^2 + b^2 + c^2 = 2abc

Then (kp)^2 + (kq)^2 + (kr)^2 = 2(kp)(kq)(kr)

k^2(p^2 + q^2 + r^2) = k^3(2pqr)

p^2 + q^2 + r^2 = k * 2pqr

...hence, the induction proof fails, since k > 1. To clarify: had we had able to prove that p^2 + q^2 + r^2 = 2pqr, then we could eliminate three even numbers as a solution, since we could divide the three number by 2 and get a second solution. Eventually, after enough divisions by 2, at least one of the numbers will be odd. However, the induction proof doesn't work, so we can't.

* * * * *

A few more random comments:

My proof and Curt's proof (below) follow similar reasoning, although his proof is more economical in that he takes care of all the powers of 2 in one fell swoop, while I work my way backwards. He's right that there has to be a maximum 2^n common factor for all three numbers, so you can jump straight to that point.

Quadrillerator's stronger statement, i.e. a^2 + b^2 + c^2 = 2^k * abc has no non-zero solutions for any k > 0, can be made even more general: In fact, there are no non-zero solutions to a^2 + b^2 + c^2 = k * abc, where k is any positive EVEN integer. The proof works like the others - prove a, b, and c must be even, then reduce it to absurdity.

That brings up a new question: what is k is odd? Playing around with this, there are non-zero solutions if k = 1 or 3. For example, one interesting solution to a^2 + b^2 + c^2 = 3abc (interesting, IMO, because a, b, and c are three different integers) is a = 1, b = 2, and c = 5. However, I can't find any solutions to a^2 + b^2 + c^2 = 5abc, or, for that matter, a^2 + b^2 + c^2 = kabc for any k > 3. If any of you can find a solution (or prove it impossible), email me - I'll post the question, and, after no one solves it, you can submit your answer to get the 10 points. :)

Answer 3

First of all, we can NOT assume a, b, c are relatively prime. Think about it.

The argument given above looking at this all modulo 4 is a good place to start, because it is indeed the case that all squares are congruent either to 0 or 1 modulo 4.

So here's the way to get a real impossibility proof (assuming they're all non-zero). Let 2^k be the LARGEST common factor of 2 among a, b, and c. k>=0

Then we can rewrite the equation as:

p^2 + q^2 + r^2 = 2^(k+1) * pqr, where p is odd.

q and r are not both odd, else the LHS is odd and the RHS is even for a contradiction. So the RHS is divisible by 4 and hence congruent to 0 mod 4.

However, p^2 is congruent to 1 mod 4, and each of q^2 and r^2 is congruent to either 0 or 1 mod 4, and so there's no way the LHS when added up is congruent to 0 mod 4.

Contradiction.

QED

So the only solution is a = b = c = 0

Answer 4

a=3, b=4 ,c=5

Answer 5

omfg, u r noob if u dont be attentive to what it is. the a2 and b2 aspects are linked (?) with the "legs" of a triangle, if it particularly is a genuine triangle. the c2 area is the hypotonuse(?). frequently, it is used to discover between the lacking aspects of a triangle.

Answer 6

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