Terrific Triplets

It is not hard to check that $a^2+b^2+c^2-ab-bc-ca$ is a factor of the expression on the left hand side. Indeed,

$a(b-c)^4 + b(c-a)^4 + c(a-b)^4$

$=(a^2+b^2+c^2-ab-bc-ca) \cdot (a^2b+ab^2+b^2c+bc^2+c^2a+ca^2-6abc).$

Let $x=b-a>0$ and $y=c-b>0$ . We have

$(x^2+y^2+xy)(ay^2+b(x+y)^2+cx^2)=836 .... (1)$

Therefore, $x^2+y^2+xy$ is a factor of $836$ . Also, if $q$ is a positive common factor of $x$ , $y$ , then by equation (1), we get that $q^4$ divides $836=2^2\cdot 11\cdot 19$ . It can only be $q=1$ . In particular, $x,y$ have different parity. So $x^2+y^2+xy$ is odd, and thus is a factor of $11\cdot 19$ .

It is also easy to check that $x^2+y^2+xy\equiv 0~or~1~(mod~3)$ . Since $11 \equiv 2~(mod~3)$ , the only possible case left is $x^2+y^2+xy=19$ . The smaller one of $x,y$ won't be greater than $2$ because $19\geq 3\cdot \min(x,y)^2$ . Therefore by symmetry it is enough to test for $x=1,2$ .

When $x=1$ , we get $y(y+1)=18$ , which has no solution; when $x=2$ , $y^2+2y+1=16=(y+1)^2$ , so $y=3$ . We get that all possible solutions are $(x,y)=(2,3)$ or $(3,2)$ . In either case, $x+y=5$ . Now $a=c-x-y=c-5$ , $b=c-y$ . From equation (1), and $x^2+y^2+xy=19$ , we get $(c-5)y^2+(c-y)\cdot 25+cx^2= 44$ $c(x^2+y^2+25) = 44 + 25y + 5y^2$ $38c=44+5y(y+5).$

When $y=2$ , $38c=114$ , so $c=3$ . When $y=3$ , $38c=164$ , which does not have integer solution for $c$ .

In conclusion, there is only one possible integer $c=3$ that satisfies the condition.

First, note that we can factorize the above expression to give

$((b-a)^2 + (c-b)^2 + (b-a)(c-b)) (a(b-c)^2 + b(c-a)^2 + c(a-b)^2) = 836$

Note that the second factor is always even. If a,b,c are all even, the expression is obviously even. If two of a,b,c are even - say a,b (without loss of generalization since the expression is cyclic), then (a-b) is even and so the whole expression is even. If only one is even - say a, then b,c,(c-a) and (a-b) are all odd, and so the first term is even while the second and third terms are odd - the resulting expression is still even. If a,b,c are all odd, (b-c), (c-a) and (a-b) are all even and so is the expression.

We now look into the first factor $((b-a)^2 + (c-b)^2 + (b-a)(c-b))$ .

If both (b-a) and (c-b) is even, then $((b-a)^2 + (c-b)^2 + (b-a)(c-b))$ is a multiple of 4. The second factor is a multiple of 2, so when combined, the expression is a multiple of 8, but 8 does not divide 836, and thus at least one of (b-a) or (c-b) is odd. If at least one of them is odd, then the resulting factor would be odd. If (b-a) is odd, then the first term is odd, while the second and third terms are even. If (c-b) is odd, then the second term is odd while the first and third terms are even. If both (b-a) and (c-b) are odd, then all terms are odd.

The factors of 836 are 1, 2, 4, 11, 19, 22, 30, 44, 76, 209, 418, 836.

Since both (b-a) and (c-b) are positive integers, the first factor must have a minimum value of 3. Also, it must be an odd integer, so we're left with 3 possibilities - 11, 19 and 209.

We let the lesser value of (b-a) and (c-b) be x and the larger value y, so we have $x^2 + y^2 + xy = 11, 19, 209 \geq 3x^2$ .

For the first case, x can only be 1. However $y^2 + y + 1 = 11$ has no integer solutions since its discriminant = 1 + 40 = 41 is not a square integer.

For the second case, x can be 1 or 2. $y^2 + y + 1 = 19$ has no integer solutions since its discriminant = 1 + 72 = 73 is not a square integer. $y^2 + 2y + 4 = 19$ has two solutions y = 3 and y = -5. Since y $>$ x, thus y = 3 is the only possibility.

For the third case, x can be 1,2,...,8. We consider the discriminants of the quadratic equation $y^2 + xy + (x^2 - 209) = 0$ , which is $x^2 - 4 (x^2 - 209) = 836 - 3x^2$ from x = 1 to 8. It can be easily seen that all such discriminants are not square integers, thus the third case fails as well.

Hence, we have $b-a=2, c-b=3$ or $b-a=3, c-b=2$ .

We consider the first case. Since the first factor is 19, the second factor is 44. Substituting b = c-3 and a = c-5, we achieve $9(c-5)+25(c-3)+4c = 44$ which gives us $38c = 164$ . c is not an integer so this case fails as well.

For the second case, we substitute b = c-2 and a = c-5, and achieve $4(c-5)+25(c-2)+9c = 44$ , which gives us $38c=114$ . This gives us $c=3$ as a solution, so $b=1$ and $a=-2$ .

Therefore, the only solution to the problem is $(a,b,c) = (-2, 1, 3)$ , so the sum of all integers c is 3.

Before you get "scared off" by the sheer length of this solution, let me assure you that most of them are motivations and case trying, so you can just skip those if you are fine with it. :)

Motivation 1 As with number theory questions, the expression can look simple, but the solution (with multiple cases) can be rather tedious. The first intuition we should have when we see such expressions is to ask ourselves if there is a way to factorise it into something more manageable. (Also refer to the tags of the question: Two of which states "Prime Factorisation" and "Factorisation". Therefore, we should be able to guess that is the way to defeat this problem.) Also, a way to go about this question is to construct variables to simplify matters. E.g. Since $a<b<c$ , let $x = b-a$ and $y=c-b$ , where $x,y >0$ . We will see more of this later in the question.

Solution Using prime factorisation on the RHS (right-hand side), we notice $836=2^2 \cdot 11 \cdot 19$ . Therefore, a way to tackle this problem will be to try out the various factors for some factorisation of the LHS.

Motivation 2 Hmm... but how can we factorise the LHS to simplify the expression? Recall Factor Theorem? (An example is as follows: Suppose a function $f(x)$ , has root $a$ such that $f(a)=0$ , then $(x-a)$ is a factor.) We can try to find the roots of the expression and from there derive possible factors!

Clearly, $a=b=c$ is a solution. (The expression equals $0$ .) Hmm... what now? Let's try to construct an expression similar to the original that could be a possible factor. Here, we conjecture that

$(b-a)^2 + (c-a)^2 + (c-b)^2 = 2(a^2 + b^2 +c^2 -ab - bc - ac)$

is a factor. Fortunately after expanding the original expression and grouping terms, we notice that the the LHS of the equation factorises into:

$a(b-c)^4 +b(c-a)^4+c(b-a)^4$ $= (a^2+b^2+c^2-ab-ac-bc)(a^2b+ab^2+b^2c+bc^2+a^2c+ac^2-6abc)$

We were right! :D

To simplify matter, let $x=(b-a)$ and $y=(c-b)$ . Note that since $a<b<c$ , $x,y>0. x,y \in \mathbb{Z}$ .

Notice the part of the factorisation: (We just need to care about this first because solving one side of the factorisation will tell us the other side automatically. E.g. If one factor is $1$ , the other must be equal to $836$ since product equals $836$ .(

$(a^2+b^2+c^2-ab-ac-bc) = x^2 +xy+y^2$ .

Here, once again be reminded that $a<b<c$ , thus $x,y >0$ . Here, we start to eliminate cases, because $836$ has $12$ factors, we wouldn't want to try all of them. :p

Notice that the factor (refer to $x^2 +xy+y^2$ ) cannot be a multiple of $2$ and not a multiple of $4$ . (In other words, all multiples of $2$ that are not divisible by $4$ are eliminated). This is because a square $\equiv 0$ or $1 \pmod 4 .$ If the expression is divisible by $2$ , the squares must be even, thus, they have the prime factor $2$ . Square $2$ , we get $4$ . Thus, all factors divisible by $2$ must be divisible by $4$ .

Case 1 $x^2 +xy+y^2 = 1$ No solution since $x,y>0$ .

Case 2 $x^2 +xy+y^2 = 836$ . No solutions, because the other factor gets too big.

Case 3 $x^2 +xy+y^2 = 4$ . Using quadratic equation, we get $y^2 - 4(y^2 - 4) = 16 - 3y^2=k^2$ , since we want $x$ to be an integer. Trying all $y$ from $1$ to $2$ , we have $y = 2$ to get $k=2$ . Sub $k=2$ back into the quadratic equation to obtain $x=\frac{-2\pm 2}{2} = -2 \text{ or } 0$ , both of which are NA (because $x>0$ .) Therefore, no solutions.

Case 4 $x^2 +xy+y^2 = 209$ . No solutions, because the other factor gets too big.

Case 5 $x^2 +xy+y^2 = 19$ . Here, $y=2, x=3$ .

Case 6 $x^2 +xy+y^2 = 44$ . No solutions, because the other factor gets too big.

Case 7 $x^2 +xy+y^2 = 76$ . Similarly, $y^2 - 4(y^2 - 76) = 304 - 3y^2=k^2$ . We get $y=4, k=16, x=6$ . However, this causes the second factor to get too big. (How do we know? Substitute into original equation, and test out values of $a,b,c$ .)

Case 8 $x^2 +xy+y^2 = 11$ . No solutions.

Thus, that the only solution is when $x^2 +xy+y^2 = 19$ and $x=3, y=2$ . Substituting this into original equation, we get $a=-2, b=1, c=3$ as our only solution. Thus, answer is $\boxed{3}$ .

Credits Last but not least, this solution is only possible with @Calvin Lin and @Eddie The Head . :)

Given conditions - $a,b,c \epsilon \mathbb{Z}$ and $a<b<c$

To find – The sum of all integer values of c in the equation $a(b-c)^4+b(c-a)^4+c(a-b)^4=836$ .

Solution - Let $a=b-x$ and $c=b+y$ where, $x,y \epsilon \mathbb {N}$ [Reason- If $x$ and $y$ are equal to zero then $a=b$ or $b=c$ and if $x$ and $y$ are negative then $a$ will be greater than $b$ , which is in contradiction to the given conditions.] Substituting for $a$ and $c$ in the equation, $(b-x)[b-(b+y)]^4 + b[b+y-(b-x)]^4 + (b+y)[b-x-b]^4 = 836$ .

On simplifying, $by^4 + by^4 + b(x+y)^4 + yx^4 -xy^4=836$ ,

$b[y^4 + x^4 +(x+y)^4] = 836 - yx^4 +xy^4$

$\therefore b= \frac {836 -yx^4 +xy^4}{[y^4 + x^4 +(x+y)^4]}$

Since, b is an integer, $[y^4 + x^4 +(x+y)^4] \mid 836 -yx^4 +xy^4$ and $836 -yx^4 +xy^4 \geq y^4 + x^4 +(x+y)^4$ ............ $(a)$ .

$\Rightarrow 836 \geq y^4 + x^4 +(x+y)^4+yx^4 -xy^4$

$\therefore 836 > (x+y)^4$

But $5^4 < 836 < 6^4$

So, $(x+y)^4 \leq 5^4$

$\Rightarrow x+y \leq 5$

Checking for all possible values of $x$ and $y$ whether they satisfy all the conditions for b as stated in $(a)$ When,

1. $x=1, y=1$ , $b=\frac{418}{9}$ .

2. $x=1, y=2$ , $b= \frac{425}{49}$ .

3. $x=2, y=1$ , $b= \frac{411}{49}$

4. $x=1, y=3$ , $b= \frac{457}{169}$

5. $x=3, y=1$ , $b= \frac{379}{169}$

6. $x=2, y=3$ , $b= \frac{25}{19}$

7. $x=3, y=2$ , $b=1$

8. $x=1, y=4$ , $b = \frac{544}{441}$

   1. $x=4, y=1$ , $b = \frac{292}{441}$ .

Values of $x$ and $y$ in $1$ to $6$ , $8$ and $9$ gives a contradictory value for $b$ because $b$ is an integer but for these values we get $b$ to be a fraction.

Therefore the only value of $x=3$ , and $y=2$ and the value $b=1$ [as calculated earliar]. Hence, there is only one value for $a$ and $c$ .

So, value of $a=b-x=1-3=-2$ and value of $c=b+y=1+2=3$ .

$\therefore$ sum of all values of c is $\boxed {3}$

a(b – c)⁴ + b(c – a)⁴ + c(a – b)⁴ = 836 with condition a < b < c.

Factorize the equation (we may recognize the form as cyclic polynomial).

(a² + b² + c² – ab – bc – ca)(a²b + a²c + ab² + b²c + ac² + bc² - 6abc) = 836

{(a – b)² + (b – c)² + (c – a)²}{a(b – c)² + b(c – a)² + c(a – b)²} = 1672

See that (a – b)² + (b – c)² + (c – a)² ≥ 0, hence 1672 is product of two positive different integer to fulfill {(a – b)² + (b – c)² + (c – a)²}{a(b – c)² + b(c – a)² + c(a – b)²}. Consider a(b – c)² + b(c – a)² + c(a – b)² is greater than (a – b)² + (b – c)² + (c – a)².

(1) {(a – b)² + (b – c)² + (c – a)²}{a(b – c)² + b(c – a)² + c(a – b)²} = 1*1672

Substitute a = –1, b = 0, c = 1 to obtain the smallest possibility of (a – b)² + (b – c)² + (c – a)². We get 5 as the result of calculation. So we may conclude that (a – b)² + (b – c)² + (c – a)² minimum at 5. By this conclusion, point (2) and (3) has no integer solution.

(2) {(a – b)² + (b – c)² + (c – a)²}{a(b – c)² + b(c – a)² + c(a – b)²} = 2*836 {look at point (1)}

(3) {(a – b)² + (b – c)² + (c – a)²}{a(b – c)² + b(c – a)² + c(a – b)²} = 4*418 {look at point (1)}

(4) {(a – b)² + (b – c)² + (c – a)²}{a(b – c)² + b(c – a)² + c(a – b)²} = 8*209

We can see that 8 is the addition of 3 square integer, 8 = (±2)² + (±2)² + 0. Pay attention that 0 only can be obtain if a = b or b = c or c = a, contradiction from condition a < b < c.

(5) {(a – b)² + (b – c)² + (c – a)²}{a(b – c)² + b(c – a)² + c(a – b)²} = 11*152

11 = (±3)² + (±1)² + (±1)².

Let a – b = b – c = ±k, c – a = ±2k. (a – b)² = (b – c)² = k², (c – a) = 4k². From here we may conclude that for two same square integers, the other integer should be divisible by 4. So point (6), (7), (8) (i) has no integer solution.

(6) {(a – b)² + (b – c)² + (c – a)²}{a(b – c)² + b(c – a)² + c(a – b)²} = 19*88

19 = (±3)² + (±3)² + (±1)² {look at point (5)}

(7) {(a – b)² + (b – c)² + (c – a)²}{a(b – c)² + b(c – a)² + c(a – b)²} = 22*76

22 = (±3)² + (±3)² + (±2)² {look at point (5)}

(8) {(a – b)² + (b – c)² + (c – a)²}{a(b – c)² + b(c – a)² + c(a – b)²} = 38*44

(i) 38 = (±6)² + (±1)² + (±1)² {look at point (5)}

(ii) 38 = (±5)² + (±3)² + (±2)²

By using trial and error, we get a = 3, b = 1, c = -2 and its permutations. Hence c = 3 (a < b < c)

I got the solution c=3 b=1 a=_2 by noting that (a-b) +(b-c) +(c-a) =0 and observing that c-a could not exceed 5 or be less than2 a little trial and error ensued but I would like to see a more rigorous solution

$I$ didn't arrive at the solution explicitly or generally. Rather, $I$ got solution after making following cases:

Case $1$ : Let $a$ , $b$ and $c$ be all negative integers such that $a<b<c$ . This results in a negative integer on left hand side, hence this case is $rejected$ .

Case $2$ : Let $a$ , $b$ be negative and $c$ be a positive integers, respectively. A look at equation reveals that the greatest integer raised to the power $four$ would be ( $c-a$ ) (assumed) or otherwise large positive integer value is required for c to make equation positive $($ a large positive integer $(Z^+)^4$ means a and b too should be large and negative justifies my assumption $)$ . Hence, this also makes equation left hand side negative, therefore $rejected$ .

Case $3$ : Let $a$ , $b$ and $c$ be all positive integers with given condition $a<b<c$ . There exists no solution for any three such positive integers. Here, ( $b-c$ ) , ( $c-a$ ) , ( $a-b) \leq 5$ . Any combination for $a$ , $b$ , $c$ where $0 \leq a \leq 3$ , $1 \leq b \leq 4$ , $2 \leq a \leq 5$ . Hence, this case is $rejected$ too.

Case $4$ : (final case) Above cases imply that only solution exists when $a$ is non-negative and $b$ , $c$ are positive integers, respectively. Here, too $I$ assumed that $(c-a)$ is the greatest term in brackets, though other could be $(c-b)$ but it is multiplied by $a$ a negative integer. Here, by hit and trial $I$ came to the following equalities and constraints :

$(c-a) = 5$ , $(a-b) = -3)$ , $(b-c) = 1$ ;

$3 \leq c \leq 4$ , $1 \leq b \leq 3$ , $-2 \leq a \leq -1$ ;

Solving above equations and inequalities give, $a$ = -2 , $b$ = 1 , $c$ = 3. So, the solutions is $c$ = 3.