Intereting Posts

Where can I find SOLUTIONS to real analysis problems?
Show that $\operatorname{diam}(A\cup B)\le \operatorname{diam}(A)+\operatorname{diam}(B)+d(A,B)$
If $a^2 + b^2 + c^2 = 2$, find the maximum of $\prod(a^5+b^5)$
Techniques for showing an ideal in $k$ is prime
Some case when the central limit theorem fails
Evaluate $\int_0^1 \frac{x^k-1}{\ln x}dx $ using high school techniques
Why does a diagonalization of a matrix B with the basis of a commuting matrix A give a block diagonal matrix?
Group Extensions
$\lim_{x\to 1^-} \sum_{k=0}^\infty \left( x^{k^2}-x^{(k+\alpha)^2}\right)$
Greatest lower bound of $\{x+k\mid x\in A\}$
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Estimate total song ('coupon') number by number of repeats
Has $S$ infinitely many nilpotent elements?
Do isomorphic structures always satisfy the same second-order sentences?
Eigenvalues for $3\times 3$ stochastic matrices

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 $$(2,2,1,1)$$

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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
,w=quadunit(8*n);[a,b]=[component(w,2),component(w,3)];if(a^2-b^2*2*n==-1,w=w^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
,w=quadunit(8*n);[a,b]=[component(w,2),component(w,3)];if(a^2-b^2*2*n==-1,w=w^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]
?
```

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