How many triplets of integers $(a, b, c)$ are there such that $a^4 + b^4 + c^4 = 2 a^2b^2 + 2 b^2c^2 + 2 c^2a^2 + 2014?$
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good solution
"BRILLIANT" Solution
exactly how i did it!
Smart way :)
Observation : The right hand side of the given equation is even. Hence the left hand side must also be even. This gives rise to two cases.
Case 1 : a,b,c are all even. In that case $a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2$ is divisible by 16, while 2014 is not. Hence this case is not possible.
Case 2: two of a,b,c are odd, the other one is even. Without any loss of generality, assume $a$ and $b$ are odd. Then it is easy to verify that $a^2=1\mod(4),b^2=1 \mod (4), c^2=0 \mod(4)$ . Hence $a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2 \mod(4)=0$ , while 2014 is not divisible by 4. Hence this case is also not possible.
Thus there does not exist any solution to this diophantine equation. $\blacksquare$
wll done meet me
Suppose there's a solution, because variable are replaceable with their additive inverse, then without the loss of the generality, we may assume that $0 \leq a \leq b \leq c$
If $a = 0$ , rearrange the equation and we get $(c^2-b^2)^2 = 2014$ , which doesn't have integer solution because $2014$ is not a perfect square. So they are all positive integers.
Because $1+1+1 \leq a^4 + b^4 + c^4 \leq 2014 < 7^4$ , we have $1 \leq a \leq b \leq c \leq 6$
Since $2$ divides $\text{RHS}$ , if we have all three of $a,b,c$ are even numbers, then $2^4$ divides $\text{LHS}$ but doesn't divide $\text{RHS}$ , which is absurd. So the triplets $(a,b,c)$ must have two odd numbers and one even number.
Noting that one of the variation of Heron's Formula is, $\text{Area } = \frac {1}{4} \sqrt{ 2\left (a^2 b^2+ a^2 c^2+b^2 c^2 \right) - \left (a^4 + b^4 + c^4 \right ) }$
If we assume $a,b,c$ are lengths of the triangle (possible degenerate triangle), then the area of the triangle is $\frac {\sqrt{-2014}}{4}$ which is invalid. So they can't be the lengths of a triangle. Thus, we have another constraint $a+b<c$
So with constraints:
$1 \leq a \leq b \leq c \leq 6$ ,
the triplets $(a,b,c)$ must have two odd numbers and one even number.
$a+b < c$
there's only a couple of trial and errors we need to carry out to determine whether they satisfy the given equation
$(1,1,4) \Rightarrow \text{ Invalid Solution}$
$(1,1,6) \Rightarrow \text{ Invalid Solution}$
$(1,2,5) \Rightarrow \text{ Invalid Solution}$
$(1,3,6) \Rightarrow \text{ Invalid Solution}$
Since we have exhausted all possible scenario, there's no integer solution to the given equation.
u made it difficult....be easy....
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THAT'S MY TALENT!
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True elegence lies in making things simpler. I can employ a really long solution based on further breaking up terms on both sides and evaluating all 4! ways. (I began doing it that way and got fed up and gave up. I tried zero and hurray!!!)
What a saying, Han
$a^{4} + b^{4} + c^{4} = 2a^{2}b^{2}+2b^{2}c^{2}+2c^{2}a^{2}+2014$ This can be written as $(a+b+c)(c-a-b)(c-a+b)(c+a-b) = 2014$ $(a+b+c)(c-a-b)(c-a+b)(c+a-b) = 2*19*53$ So one of the 4 terms is even. But we see that $(a+b+c)-(c+a+b)) = 2(a+b)$ and $(c-a+b)-(c+a-b) = 2(a-b)$ . So if one is even the other term must also be even but two terms cannot be even as 2014 contains only one 2 in it's prime factors....Hence no solutions are possible....
It doesn't take long to notice that bringing all the terms, except 2014, to the left-hand-side, the expression can be factored as:
$(a^2-b^2-c^2)^2=2014$ .
We make this observation by noting that $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)$ . Similarly, if we have $-2ab,-2bc,-2ca$ instead, the factorisation simply becomes $(a-b-c)^2$ .
From this observation, we take the square-roots of both sides and bring $c^2$ over to the right-hand-side:
$\implies a^2-b^2 = c^2 \pm \sqrt{2014}$ .
Now note that $a$ and $b$ are both integers hence the L.H.S is always an integer. On the R.H.S, $c^2$ is always an integer but $\sqrt{2014} \notin \mathbb{Z}$ , therefore the R.H.S is never an integer. Hence the number of solutions to this equation is trivially $\boxed{0}.$
muhammad shariq , i think you made a wrong assumption there. (a^2-b^2-C^2)^2 is not correct do it again
at the very first line of your solution u can see that 2014 is not a perfect square so there can't be such combination of integers............................
that was tricky, dunno why I had to think hard. Thats why, thinking with heart gives a good result :P
[; (a^2+b^2+c^2)=2(2(a^2b^2+b^2c^2+a^2c^2)+1007) ;]
The right hand side is even but doesn't have [; 4 ;] as a factor, so is never a square.
the equation is equivalent to: (a-c-b)(a+c+b)(a-c+b)(a+c-b)=2x53x19
Dnt knw why I can't solve the problems.maybe I am very young. In class 9.
$a^{4} + b^{4} + c^{4} - 2a^{2}b^{2} - 2b^{2}c^{2} - 2c^{2}a^{2} = 2014$
$(a^{2} - b^{2} - c^{2})^{2} - 4b^{2}c^{2} = 2014$
$(a^{2} - b^{2} - c^{2} - 2bc)(a^{2} - b^{2} - c^{2} + 2bc) = \sqrt{2014}$
Since a, b and c are integers, then $a^{2} - b^{2} - c^{2} - 2bc$ and $a^{2}-b^{2}-c^{2} + 2bc$ are integers. But since $\sqrt{2014}$ isn't an integer. There's no solution for it.
Please check your second step, the squaring formula is incorrect.
in second step subtract 4(b^2)(c^2).
Thank you so much for reminding me. This is a stupid mistake and I will try to minimalize it.
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$a^4 + b^4 + c^4 - 2a^2b^2 - 2b^2c^2 - 2a^2c^2 = 2014$
$(a^2 + b^2)^2 - 4a^2b^2 + c^4 -2c^2(a^2 + b^2) = 2014$
$(a^2 + b^2 -c^2)^2 - 4a^2b^2 = 2014$
$(a^2 + b^2 - c^2 - 2ab)(a^2 + b^2 -c^2 + 2ab) = 2014$
Suppose there are integer solutions.
Now, since $2014 = 2 \times 1007$ , one and only one of the terms must be divisible by $2$ . But since the terms differ by $4ab$ , if one term is divisible by 2, the other term MUST be divisble by 2, which leads to a contradiction. Therefore, there exist no integer solutions and the answer is $\boxed{0}$