# Equation over the integers

Find all quartuples $(a,b,c,d)$ with non-negative integers $a,b,c,d$ satisfying $$2^a3^b-5^c7^d=1$$

One solution is $$(2,2,1,1)$$

#### Solutions Collecting From Web of "Equation over the integers"

This is more of a heuristic since it makes use of the proven / unproven abc-conjecture:

Let $x = 5^c7^d,y = 1, z = 2^a3^b$ then $z = x+y$. Assume that the abc-conjecture holds for $\epsilon = 1, K_{\epsilon} = 1$ then we have:
$z < rad(xyz)^2$ from which it follows that $2^a 3^b < 2^2 3^2 5^2 7^2$ and since $5^c7^d < 2^a 3^b < 2^2 3^2 5^2 7^2 = 44100$ we see that there are at most finitely many $(a,b,c,d)$. Solving $2^a = 44100$ we see that $a \equiv 15$. Hence we can search for in the range $0\le a,b,c,d \le 16$ which gives the solutions Peter mentioned.

I just remembered a possibility to find all solutions. The key is the theorem of stormer (look at wikipedia for details). It allows to calculate all pairs of numbers with differnce $1$ only having prime factors contained in a given list.

For the list $[2,3,5,7]$, we get the following pairs (calculation was done in PARI/GP) :

? p=[2,3,5,7];s=max(3,(vecmax(p)+1)/2);fordiv(prod(j=1,length(p),p[j]),n,if(n<>2
;for(j=1,s,a=component(w^j,2);[u,v]=[(a-1)/2,(a+1)/2];[q,r]=[u,v];for(j=1,length
(p),while(Mod(q,p[j])==0,q=q/p[j]);while(Mod(r,p[j])==0,r=r/p[j]));if(q*r==1,pri
nt([u,v])))))
[1, 2]
[8, 9]
[49, 50]
[2, 3]
[24, 25]
[2400, 2401]
[9, 10]
[3, 4]
[48, 49]
[7, 8]
[224, 225]
[4, 5]
[80, 81]
[63, 64]
[5, 6]
[6, 7]
[4374, 4375]
[15, 16]
[125, 126]
[27, 28]
[35, 36]
[14, 15]
[20, 21]
?


You can check which pairs have the required factorizations (namely the first number only $5’s$ and $7’s$, and the second only $2’s$ and $3’s$. It will turn out that we only have $4$ such pairs. So the list above is actually complete.

Update : I slightly improved the output, showing the prime factors of the numbers :

? p=[2,3,5,7];s=max(3,(vecmax(p)+1)/2);fordiv(prod(j=1,length(p),p[j]),n,if(n<>2
;for(j=1,s,a=component(w^j,2);[u,v]=[(a-1)/2,(a+1)/2];[q,r]=[u,v];for(j=1,length
(p),while(Mod(q,p[j])==0,q=q/p[j]);while(Mod(r,p[j])==0,r=r/p[j]));if(q*r==1,pri
nt([u,v],"   ",component(factorint(u),1)~,"   ",component(factor(v),1)~)))))
[1, 2]   []   [2]
[8, 9]   [2]   [3]
[49, 50]   [7]   [2, 5]
[2, 3]   [2]   [3]
[24, 25]   [2, 3]   [5]
[2400, 2401]   [2, 3, 5]   [7]
[9, 10]   [3]   [2, 5]
[3, 4]   [3]   [2]
[48, 49]   [2, 3]   [7]
[7, 8]   [7]   [2]
[224, 225]   [2, 7]   [3, 5]
[4, 5]   [2]   [5]
[80, 81]   [2, 5]   [3]
[63, 64]   [3, 7]   [2]
[5, 6]   [5]   [2, 3]
[6, 7]   [2, 3]   [7]
[4374, 4375]   [2, 3]   [5, 7]
[15, 16]   [3, 5]   [2]
[125, 126]   [5]   [2, 3, 7]
[27, 28]   [3]   [2, 7]
[35, 36]   [5, 7]   [2, 3]
[14, 15]   [2, 7]   [3, 5]
[20, 21]   [2, 5]   [3, 7]
?