My calculator says it's true

I have discovered a truly marvellous proof of this, which this margin is too narrow to contain.

This is a pretty funny problem, actually. Obviously, it is not true, because it is a violation of Fermat's Last Theorem. This is known as a "Fermat near miss." A calculator will tell you that two sides are equal, but there are digits of the numbers after what the calculator will tell you that are not equal.

Another way to tell that the two sides aren't equal is to take both sides $\text{mod}$ easy to work with numbers. Taking each side $\text{mod }2$ and $\text{mod }10$ is inconclusive. The left side, if it is in fact an integer, would be divisible by $3,$ because $3987^{12}$ and $4365^{12}$ are both divisible by $3^{12}.$ However, $4472$ is not divisible by $3,$ so $4472^{12}$ is not divisible by $3$ either.

You can learn more about Fermat near misses in this video from Numberphile.

Assuming it to be true

3987^12 = 4472^12-4365^12 = (4472-4365)(.........) = 107 * (..)

Since 3987 is not divisible by 107, it has to be false

A violation of Fermat's Last Theorem

This is known as Homer's Last Theorem from the Simpsons (the episode was "Wizard of Evergreen Terrace"

The equation above and the inequality below are also quite interesting...

The LHS is clearly divisible by 3 but the RHS is not !! Buy a better calculator mate!! :D :D

9 is a common factor for both 3987 and 4365.. taking the common factor out of the root, implies that 4472 should be divisible by 9.. but it is not.. hence the given equation doesn't hold good.

$\sqrt[12]{3987^{12} + 4365^{12}} = 4472$ ,

then $3987^{12} + 4365^{12} = 4472^{12}$

Or by Fermat's last theorem : there is no three positive integers $x$ , $y$ , $z$ , that satisfy $x^n + y^n = z^n$

if true, then 3987^12+4365^12=4472^12, and obviously is a violation of Fermat's theorem

According to Fermat's Last Theorem, for integers $n > 2$ , there are no integers $a$ , $b$ , and $c$ for which $a^n + b^n = c^n$ . Since, $12 > 2$ , there cannot be integers $a$ , $b$ , and $c$ such that $a^{12} + b^{12} = c^{12}$ .

Therefore, there is no integer $c$ where $3987^{12} + 4365^{12} = c^{12}$ , meaning that there is no integer $c$ such that $\sqrt[12]{3987^{12} + 4365^{12}} = c$ . Thus the value of $\sqrt[12]{3987^{12} + 4365^{12}}$ is not an integer, while the above equation asserts that $\sqrt[12]{3987^{12} + 4365^{12}}$ is equal to the integer $4472$ , meaning that the above statement is false .

Since it's just asked to check the correctness we could check by considering the common factors on both the side. Now for LHS Nos, we get 3987 = 1329 * 3 and 4365= 1455*3, raising to power 12 and taking 3^12 common LHS becomes multiple of 3 where RHS which is 4472 isnt. SO it must be false. Simple...!

fermat's last theorem

Actually the value of: $({3987^{12} + 4365^{12}})^{1/12} = 4472.0000000070592 (approx.)$

Thus, $({3987^{12} + 4365^{12}})^{1/12}$ is not equal to $4472$ .

So, the answer is $\boxed{No}$ .

Use logarithm, and solve with appoxrimate values.

Easiest way to think is LHS is divisible by 3, but RHS is not

I did this very basically. Not applying any fermat principles here. from 4365 I wrote it as 3987+378 so it becomes 12th root of:- 3987^12+(3987+378)^12 Now from the second bracket too 3987 common to give, 3987^12+3987^12(1+0.094), ignoring 0.094, just for convinience, we get 3987 12th root of 2 which is 3987 1.05 which is 4186 which led me to answer. Of course I had seen the Numberphile video but this was just some simple method I used. I do agree its not the best of mathematical methods as difference between idealized answer and calculator answer are not very far but I reasoned that 0.094 will not have affected 12th root much. I Can be wrong here though, correct me as needed.

this is not (a+b)^2 type to take out a+b from root, it is a^2 + b^2 at any cost find out a^12 value and b^12 value add them then go for 2x2x......... from that take out one number of 12 same nos. if it is perfect 12th root we will get nos. without remain odd multiply selected nos result is answer

$2\ 3^{24}\ 17\ 11593\ 32353\ 85313\ 208212548830074457$ , not even close.

It's because the calculator can't show all the digits so it truncates all but the last few digits also FLT

Because no number greater than 2 exist which satisfy x^n+y^n=z^n

If $x^n + y^n = z^n, n > 2, n \in \mathbb{Z}, xyz \neq 0$ , then there are no solutions for $(x,y,z)$ such that $x,y,z \in \mathbb{Z}$ .

Using the divisibility test, 3987 and 4365 are both divisible by 9, while 4472 is not. For 4472 to be a solution, it would need to be divisible by gcd(3987,4365)

As seen on the Simpsons, Homer Simpson brilliant "solved" FLT, however this was later proven false by Andrew Wiles.

FERMATS LAST THEOREM......

it cannot be true because of Fermat's last theorem. x^n+y^n=z^n if n>2 and n is positive integer have no whole number solution

Hint: Use The Even-Odd theory to justify the answer.

Just check modulo 3 : 3987 mod 3 = 4365 mod 3 = 0 but 4472 mod 3 = -1 therefore 0 would be equal to 1 !

From Fermat's Last Theorem x^12 + y^12 not equal to z^12.. So it's always False.

One possible solution is to determine if the two numbers on the sides of the equations, that is after raising both sides to 12, will be both divisible by a certain number. In this case, I tried 9.

Note that $3987 \pmod{9} \equiv 0$ while $4365 \pmod {9} \equiv 0$ . Thus, $3987^{12} + 4365^{12} \pmod{9} \equiv 0$

On the other hand, $4472 \pmod {9} \equiv -1$ . Thus, $4472^{12} \pmod {9} \equiv 1$ . Therefore, they are not equal.

i try to use simple number, like 4, 3 , and 2.

change 12 becomes 3, and 3987 becomes 4 and 4365 becomes 2 (and also try another numbers)

and the result if you change the number is 4,16.

so the conclusion is that the biggest number is the root is the answer

(3987,13)=1 (4365,13)=1 Following Fermat'sTheorem,we have 3987^12=1(mod 13) and 4365^12=1(mod 13) In the other hand, 4472=0(mod 13) So, the following equation is false.

It can be rewritten as follows:

$3987^{12} + 4365^{12} = 4472^{12}$

This is the same that

$9^{12}(443^{12} + 485^{12}) = 8^{12}13^{12}43^{12}$

The left hand side is divisible by $3$ while the right side is not. Therefore the answer is NO .

Let have a look at this equation if it is true (actually it isn’t). √(12&〖3987〗^12+ 〖4365〗^12 )=4472  〖3987〗^12+〖4365〗^12=〖4472〗^12  〖3987〗^12= 〖4472〗^12-〖4365〗^12 * We have: 〖4472〗^12-〖4365〗^12=(〖4472〗^6-〖4365〗^6 )(〖4472〗^6+〖4365〗^6) 〖4472〗^6-〖4365〗^6=(〖4472〗^3-〖4365〗^3 )(〖4472〗^3+〖4365〗^3 ) 〖4472〗^3-〖4365〗^3=(4472-4365)(〖4472〗^2+4472×4365+〖4365〗^2 )=107(〖4472〗^2+4472×4365+〖4365〗^2 ) If * is true, 3987/107 is an integer. Howerver it is not an integer. Thus, the equation is wrong. P/s: I don't know why it doesn't have its origin. So sorry.

No simple bcz both sides are not divisible by 3 so they are not equal also 12 is cut by 12underroot and both sum is not equal to right hand side

Note, 3 divides 3987 and 4365 but it does not divide 4472. Thus, we have a contradiction.