Prove that there is only one one pair of integers (a ,b) satisfying the equation
a-b=2019.

First - the single solution is 2019 = 45^2 - 3!

This is the result of a 15 line program computer search shown below.

Now, the proof:

Suppose both a and b are factorials. Proof that this is impossible: 

Factorials have as their last digit: 0, 1, 2, 4, or 6 

To get a subtraction to yield as its last digit 9, we have to have:

I.   a ends in 0 and b ends in 1

II.  a ends in 1 and b ends in 2 

I. This is impossible since 2020 is not a factorial.

II. Has no candidates. (only 0!, 1!, 2! have either attribute.) 

So a is a square.

A square (other than 0) ends in 1, 4, 5, 6, or 9.

Thus, the possible a, b last digit pairs that deliver a one's digit 9 are:

1 - 2

5 - 6

9 - 0

The only factorial that ends in 2 is 2! and 2021 is not a square. 

As for the 9 - 0 option, there are few factorials ending in ...#0, where # is not zero. If we look for a a square that ends in 19, we find there are none. ....ij^2 can not yield ...19.

That leaves the only factorial that ends in 6, 3!, and 2019 + 3! is a square, as above.

QED

------------

        program wq

        af=1

        do i=1,1000

        aa=i**2

        if(i.lt.50)af=af*i

           bf=1

           do j=1,50

           bf=bf*j

           dif1=aa-bf

           dif2=af-bf

           if(i.le.50.and.dif1.eq.2019)print*,'a^2 b',aa,j

           if(dif2.eq.2019)print*,' a b ',i,j

           enddo

        enddo

        end

rabbit-3:~ lord$ wq

 a^2 b    2025.00000               3

Edited on April 11, 2019, 3:27 pm

Posted by Steven Lord on 2019-04-10 15:31:45