Do miners get the transaction fee in a UTXO or as the difference between total input & total output?
-I recall from MIT cryptocurrency lectures they say the difference in value goes to miners as a fee.
-Then I read this paper counting a fee UTXO out of every transaction (they are building equations according to this)
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So is it possible both ways?or do they just have to subtract 1 from each EQ?
(It doesn't destroy their results, it's just they're writing it in clear words several times)
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Ps
Sorry if the question is too simple for the group, I'm kind of a researcher not a real user
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Thanks

None of this answer my simple question:
The fee is paid implicitly as the difference bet input &output?
- or through a UTXO containing it?
-Or both options possible?

First of all: the way you phrased our question was everything but simple.
Secondly: the answer you received from Husires was a perfect explanation.
Take your time to read it through again, and you'll understand

A block 'pays' the base reward
 +
'the difference between total input & total output' of all transactions confirmed
This means all transactions in the block excluding the coinbase transaction.

How it pays that is up to the 'miner' or 'pool'
The coinbase (first) transaction output can send it all to one address, or break it up to 2 or more addresses,
just the total output must match the total block reward.

It typically goes to one or two outputs.
Some miners cannot handle a coinbase with many outputs, so this tends to promote pools to only use a small number of outputs.
Also, of course, if the coinbase has a large number of outputs, this will mean a larger block to distribute around the network, since the coinbase transaction is always unknown to any other miner/pool on the network before the block is found.

How are these different from one another? they both lead to the same answer, let's assume a block with 2 transactions

input 1 = 10
input 2 = 15
output 1 =5
putout 2= 10

First method :

fees = (input 1 + input 2) - (output 1 + output 2 )
fees = (10 + 15) - (5 + 10 ) = 10

Second method

fees = (input 1 - output 1) + (input 2 - output 2)
fees = (10 - 5 )  + (15 - 10) = 10


As far as speed is concerned, getting the SUM of inputs and outputs and performing a single subtraction would most likely be a lot more effienct than performing n number of subtraction and then getting the SUM of all of them, but will the number differ? NO.

Now, on how the miner gets the block fees, they get it as an output that has no input, if the total difference between inputs and outputs was 1BTC and the block reward was 6.25BTC a new output will be created out of "nowhere" with a total value of 7.25BTC, if the coinbase transaction contains only 1 address the whole 7.25 will go to that address and if it contains 500 addresses it will go to all of them according to how the miner actually wants.

So to sum up, the fees themselves are NOT outputs but when the block is mined the fees become a new output spendable by the miner.

Turn on the solo mode of your Bitcoin core, and then connect all your ASIC devices to your Bitcoin core. You will have the opportunity to receive block rewards and transaction fees. 1Ph takes about one month.

1-The question discusses a paper about Bitcoin not altcoin,
 
2-I tried to end the thread by stating in each reply that I did find the answer or u may say asserted my belief that the paper was wrong in saying there is a fee UTXO for every Bitcoin transaction, but some still answering without reading the question.
3-I'm not trying to be aggressive in the previous reply, I honestly think the commenter probably intended to put that comment in another post.