Easter Dates Different

Question

Show that there in no natural number $n$ such that, for each year $A$, the years $A$ and $A+n$ have the Easter with the same date. Use a computer program for your proof: A number of programs of Easter Date have been set up, but to make it simple, use the following which is applicable for calculators.

Input "YEAR", A
A+1-19(int(A/19))->G
int(A/100)+1->C
int(3*C/4)-12->X
int((8*C+5)/25)-5->Z
int(5*A/4)-X-10->D
11*G+20+Z-X->B
B-30(int(B/30))->E
If ((E=25) and (G>11)) or (E=24)
  E+1->E
44-E->N
If N<21
  N+30->N
D+N-7*int((D+N)/7)->F
N+7-F->M
If M>31
Then
  Disp M-31, "APRIL"
Else
  Disp M, "MARCH"
END

Answer 1

A computer program cannot prove the non-existence of an entity.

Answer 2

Actually I gave this some more thought and testing:

every 400 years the dates actually do repeat perfectly.

Now if you take the exact revolution of the moon into account, it would be:

N = days it takes the moon to orbit the earth.

Find a * 400 so that N divides this PERFECTLY (with no rest), and you have a constant n after which easter will repeat itself.

Answer 3

My understanding of the question

I'm not entirely sure how to interpret your question, but I assume taht you take the program you quoted as your definition of the Easter date, and are asking whether or not this program will result in the same answer for $A$ and $A+n$ for a fixed and suitably chosen $n$ and for all years $A$. In other words:

$$ \exists n\in\mathbb N\;\forall A\in\mathbb N:f(A) = f(A+n)$$

where $f$ denotes the function $A\mapsto M$ described by your program code, without any further relation to astronomical aspects which lead to this program.

There is such an n

If this is the question, then the answer is that there is such an $n$, even though your question seems to imply the opposite. The smallest $n$ satisfying your requirements would be the following:

$$ n = 100\cdot4\cdot25\cdot19\cdot30 = 5{,}700{,}000 $$

Proof for the existence of a period

The argument goes as follows:

• Increasing $A$ by $19$ will result in the same value of $G$
• Increasing $A$ by $100$ will increment $C$ by a fixed constant
• Increasing $A$ by $100\cdot4$ will increment $X$ by a fixed constant
• Increasing $A$ by $100\cdot25$ will increment $Z$ by a fixed constant
• Increasing $A$ by $100\cdot4$ will increment $D$ by a fixed constant
• Increasing $A$ by $100\cdot4\cdot25\cdot19$ will increment $B$ by a fixed constant
• Increasing $A$ by $100\cdot4\cdot25\cdot19\cdot30$ will result in the same value of $E$, both before and after the If
• The same holds for $N$
• Increasing $A$ by $100\cdot4\cdot25\cdot19\cdot30$ will increment $D+N$ by a fixed constant (see below for the value of that constant)
• Increasing $A$ by $100\cdot4\cdot25\cdot19\cdot30\cdot7 = 7n$ will result in the same value for $F$
• If $N$ and $F$ are the same, then so is $M$ and hence the date of Easter

So the above will give a possible period:

$$7n=100\cdot4\cdot25\cdot19\cdot30\cdot7=39{,}900{,}000$$

Finding the shortest period

The period found in this way is seven times larger than the shortest possible period. The shortest possible period must be a divisor of the period already found. So when searching for a shorter period, concentrating on divisors helps reducing the search space, and prime factor decomposition will be a valuable tool.

The actual shortest period can be found with high probability using a computer experiment: Stripping one prime factor at a time, one can check whether the resulting number still can be a valid period, simply by comparing the results for a couple of years. For all prime factors except $7$, this fails quickly, therefore proving that the corresponding prime factor has to be present in the shortest possible period. On the other hand, when removing the unneccessary factor of seven, one will observe that for many years, the numbers will still agree.

Of course that is no proof that they will be equal for all years. To obtain such a proof, one can compute the change to $D+N$ if $A$ increases by $n$. That number is fixed, as argued above, so it can be computed for any combination of $A$ and $A+n$. The resulting difference turns out to be $7{,}082{,}250=7\cdot1{,}011{,}750$. Since this is a multiple of $7$, the value $F$ which depends on $(D+N)\bmod7$ will already repeat with a lag of $n$ years, and not only after $7n$ years.