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Solve $2^a=3^b5^c+7^d$ over the positive integer.

I know $a$ is even because: $(-1)^a \equiv2^a = 3^b5^c+7^d \equiv1 \ (mod\ 3)$

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You mean $a$ is even. Inspection is enough to find a solution. What values have you tried? –  Michael Biro Feb 9 '13 at 8:03

3 Answers 3

up vote 7 down vote accepted

Your title says $\in \mathbb{Z}$ but your question says positive integers. Which is it?

I will solve assuming positive integers below. If you want over $\mathbb{Z}$, I can edit it in later.

Taking $\pmod 3$ gives $2^a \equiv 1 \pmod 3$ so $2 \mid a$.

Taking mod $5$ gives $2^a \equiv 7^d \equiv 2^d \pmod 5$ so $a \equiv d \pmod 4$.

In particular, $d$ is even, so let $a=2a_1, d=2d_1, a_1 \equiv d_1 \pmod 2$.

Then $3^b5^c=(2^{a_1}-7^{d_1})(2^{a_1}+7^{d_1}), \, \gcd((2^{a_1}-7^{d_1}), (2^{a_1}+7^{d_1}))=\gcd((2^{a_1}-7^{d_1}), 2^{a_1+1})=1$

Consider 2 cases: $2 \mid {a_1, d_1}$, $2 \nmid {a_1, d_1}$

Case 1: $2 \mid {a_1, d_1}$

Then $2^{a_1}+7^{d_1} \equiv 2 \pmod 3, 2^{a_1}+7^{d_1}>1$, so $2^{a_1}+7^{d_1}=5^c, 2^{a_1}-7^{d_1}=3^b$.

Write $a_1=2a_2, d_1=2d_2$, so $3^b=(2^{a_2}-7^{d_2})(2^{a_2}+7^{d_2}), \, \gcd((2^{a_2}-7^{d_2}),(2^{a_2}+7^{d_2}))=1$.

Thus $2^{a_2}-7^{d_2}=1, 2^{a_2}+7^{d_2}=3^b$.

Taking $\pmod 7$, $2^{a_2} \equiv 1 \pmod 7$, so $3 \mid {a_2}$.

Let $a_2=3k$, then $7^{d_2}=2^{3k}-1=(2^k-1)(2^2k+2^k+1)$.

Since $\gcd((2^k-1),(2^2k+2^k+1)) \mid 3$, $\gcd((2^k-1),(2^2k+2^k+1))=1$, so $2^k-1=1$.

Thus $k=1$, so $a_2=3, d_2=1, 2^{a_2}+7^{d_2}=15 \not =3^b$, a contradiction.

Case 2: $2 \nmid {a_1, d_1}$

Then $2^{a_1}+7^{d_1} \equiv 0 \pmod 3$, so we have 2 sub cases: $2^{a_1}+7^{d_1}=3^b5^c$, $2^{a_1}+7^{d_1}=3^b$

Case 2a: $2^{a_1}+7^{d_1}=3^b5^c$.

Then $2^{a_1}-7^{d_1}=1$, and the argument as above (taking $\pmod 7$ then factoring the difference of cubes) shows $a_1=3, d_1=1$, so $2^{a_1}+7^{d_1}=15=3^b5^c$.

Thus $b=c=1$, and $a=6, d=2$, giving the solution $(a, b, c, d)=(6, 1, 1, 2)$, which works.

Case 2b: $2^{a_1}+7^{d_1}=3^b$.

Then $2^{a_1}-7^{d_1}=5^c$

Taking $\pmod 7$ gives $2^{a_1} \equiv 3^b \pmod 7$.

Since $2$ is a quadratic residue $\pmod 7$, but 3 is not a quadratic residue $\mod 7$, it follows that $2 \mid b$

Thus $3^b \equiv 1 \pmod 4$, so $2^{a_1+1}=3^b+5^c \equiv 2 \pmod 4$, a contradiction.

In conclusion $(a, b, c, d)=(6, 1, 1, 2)$ is the only positive integer solution.

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One solution to your equation is $a = 2$, $b = 0$, $c = 0$ and $d = 0$. Another one would be $a = 3$, $b = 0$, $c = 0$, $d = 1$. Yet another one would be $a = 4$, $b =1$, $c = 1$, $d = 0$. And $a = 5$, $b = 0$, $c = 2$, $d =1$.

One is clever to recognize that $a + b + c+d = 2(a - 1)$ for all $a,b,c,d \in \mathbb{N}$ that satisfy your equation. Since we want positive integer solutions, it turns out that there is only one tuple which satisfies the above: $(6,1,1,2)$.

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There are other trivial solutions, which do not satisfy your relation: $a=1,b=c=d=0$, $a=2,b=1,c=d=0$. The congruence is invalid if $b\neq 0$: in that case the second factor satisfies $3^b5^c\equiv (-1)^c\pmod 3$. –  Andrea Orta Feb 9 '13 at 9:04
@AndreaOrta I pointed out that the solutions have the above statement true, not that it is true for all $a,b,c,d$. I have explicitly said that it is true for all $a,b,c,d$ that satisfy the equation. Anyway, you are correct. –  Parth Kohli Feb 9 '13 at 9:12
Well, (2,1,0,0) is a solution and doesn't satisfy your equation, am I wrong? :) –  Andrea Orta Feb 9 '13 at 9:19
Oh, whoops... You are correct. –  Parth Kohli Feb 9 '13 at 9:29
@IvanLoh got it right! –  Parth Kohli Feb 9 '13 at 9:38

One solution in positive integers is $a=6$, $b=c=1$, $d=2$. A computer search may reveal others.

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But how to find that ? –  Xeing Feb 9 '13 at 8:30
One one decides $a$ and $d$ should be even, this example is really the first candidate. –  André Nicolas Feb 9 '13 at 8:40

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