# Find all positive integers $a, b, c$ such that $21^a+ 28^b= 35^c$ .

Find all positive integers $$a, b, c$$ such that $$21^a+ 28^b= 35^c$$.

It is clear that the equation can be rewritten as follows: $$(3 \times 7)^a+(4 \times 7)^b=(5 \times 7)^c$$ If $$a=b=c=2$$ then this is the first possible answer to this issue. It is also obvious that the sum of $$(3*7)^a+(4*7)^b$$ must end and be divisible by $$5$$. Since $$21^a$$ always ends at $$1$$, then $$28^b$$ should end at $$4$$ . Defined $$b$$ as $$b=2+4k$$ -- even positive integer.

• How about Fermat's Last Theorem? – Toby Mak Mar 6 '19 at 12:37
• Two of the three values must be equal. Say $a=b$. Consider an odd prime $p$ dividing $m$. Then $9^p+16^p$ must have only 5,7 as prime factors. – Aravind Mar 6 '19 at 13:25
• What about it? @TobyMak – Aqua Mar 6 '19 at 13:45
• @TobyMak How does Fermat's Last Theorem apply to this? The exponents don't have to be the same. – John Douma Mar 6 '19 at 13:46
• See also this similar question. – Fabio Lucchini Mar 6 '19 at 14:12

Note that $$21=3\times7$$, $$28=4\times7$$ and $$35=5\times7$$, and so by unique factorization the numbers $$21^a$$, $$28^b$$ and $$35^c$$ are all distinct for all positive integers $$a$$, $$b$$ and $$c$$. By unique factorization we see that the left hand side of $$21^a+28^b=35^c,$$ is divisible by $$7^{\min\{a,b\}}$$ and hence $$c\geq\min\{a,b\}$$. Moreover the right hand sides of $$21^a=35^c-28^b \qquad\text{ and }\qquad 28^b=35^c-21^a,$$ are divisible by $$7^{\min\{b,c\}}$$ and $$7^{\min\{a,c\}}$$, respectively, because $$28^b\neq35^c\neq21^a$$. This implies $$a\geq\min\{b,c\} \qquad\text{ and }\qquad b\geq\min\{a,c\},$$ from which it follows that $$a=b=c$$. Dividing out the factor $$7^a$$ leaves us with $$3^a+4^a=5^a,$$ which clearly has the unique solution $$a=2$$.

Reducing the equation mod $$5$$ and $$8$$ shows that $$b\equiv2\pmod{4}$$ and $$a$$ and $$c$$ are even, because $$1^a+3^b\equiv0^c\pmod{5},$$ $$5^a+4^b\equiv3^c\pmod{8}.$$ Let $$a':=\tfrac{a}{2}$$, $$b':=\tfrac{b}{2}$$ and $$c':=\tfrac{c}{2}$$ and $$d:=\min\{a',b',c'\}$$. Then we have a Pythagorean triple $$\left(21^{a'}\right)^2+\left(28^{b'}\right)^2=\left(35^{c'}\right)^2,$$ with $$\gcd\left(21^{a'},28^{b'},35^{c'}\right)=7^d$$. This means there exist coprime integers $$m$$ and $$n$$ such that $$21^{a'}=7^d(m^2-n^2),\qquad 28^{b'}=7^d(2mn),\qquad 35^{c'}=7^d(m^2+n^2),\tag{\ast}$$ and without loss of generality $$m>n>0$$. The latter identity shows that $$5^{c'}7^{c'-d}\equiv m^2+n^2\not\equiv0\pmod{7},$$ and so $$d=c'$$. It follows that if $$b'=1$$, then $$1=b'=d=c'$$ and hence $$21^a+28^2=35^2,$$ which shows that $$a=b=c=2$$. Now suppose toward a contradiction that $$b'>1$$.

The middle identity in $$(\ast)$$ shows that either $$(m,n)=(7^{b'-d},2^{2b'-1}) \qquad\text{ or }\qquad (m,n)=(2^{2b'-1},7^{b'-d}),$$ and in either case, plugging this into the last identity in $$(\ast)$$ shows that $$5^{c'}=2^{4b'-2}+7^{2b'-2d},$$ where we used that $$c'=d$$ to cancel the factors $$7^{c'}$$ and $$7^d$$. Reducing mod $$8$$ and mod $$5$$ shows that $$\begin{eqnarray*} 5^{c'}&\equiv&4^{2b'-1}+1^{b'-d}&\equiv&1&\pmod{8},\\ 0^{c'}&\equiv&4^{2b'-1}+4^{b'-d}&\equiv&(-1)+(-1)^{b'-d}&\pmod{5}, \end{eqnarray*}$$ which shows that $$c'$$ is even, say $$c'=2c''$$, and that $$b'\equiv d\pmod{2}$$. Then $$b'$$ is also even, contradicting the fact that, $$b\equiv2\pmod{4}$$. Hence $$a=b=c=2$$ is the unique solution.

Note that $$21=3\times7$$, $$28=4\times 7$$ and $$35=5\times7$$ are all multiples of $$7$$. Suppose $$b>a$$. Then $$5^c7^c=3^a7^a+4^b7^b=(3^a+4^b7^{b-a})7^a,$$ where $$7^{b-a}$$ is an integer divisible by $$7$$, and so $$3^a+4^b7^{b-a}$$ is not divisible by $$7$$. By unique factorization it follows that $$a=c$$ and so $$5^a=3^a+4^b7^{b-a}.\tag{1}$$ Reducing this identity mod $$4$$ and mod $$5$$ shows that $$a$$ and $$b$$ are even, respectively, because $$\begin{eqnarray*} 1^a&\equiv&3^a+0^b3^{b-a}&\equiv&3^a&\pmod{4},\\ 0^a&\equiv&3^a+4^b2^{b-a}&\equiv&3^a(1+3^{b})&\pmod{5}. \end{eqnarray*}$$ Let $$a':=\tfrac{a}{2}$$ and $$b':=\tfrac{b}{2}$$. Then from $$(1)$$ we find that the primitive Pythagorean triple $$\left(3^{a'}\right)^2+\left(4^{b'}7^{b'-a'}\right)^2 =\left(5^{a'}\right)^2,$$ which means there exist coprime integers $$m$$ and $$n$$ such that $$3^{a'}=m^2-n^2,\qquad 4^{b'}7^{b'-a'}=2mn,\qquad 5^{a'}=m^2+n^2,\tag{2}$$ and without loss of generality $$m>n>0$$. The middle identity shows that either $$(m,n)=(7^{b'-a'},2^{2b'-1}) \qquad\text{ or }\qquad (m,n)=(2^{2b'-1},7^{b'-a'}),$$ but in the latter case we get $$b=2$$ because $$3^{a'}=2^{2b-2}-7^{b-a}\equiv4^{b-1}-1\pmod{8},$$ and $$a$$ and $$b$$ are even; but $$b>a>0$$, a contradiction. Hence $$m=7^{b'-a'}$$ and $$n=2^{2b'-1}$$, and plugging these back into $$(2)$$ shows that $$3^{a'}=7^{b-a}-2^{2b-2}\qquad\text{ and }\qquad 5^{a'}=7^{b-a}+2^{2b-2},$$ and hence $$5^{a'}-3^{a'}=2^{2b-1}$$. Reducing mod $$8$$ shows that $$a'$$ is even, say $$a'=2a''$$, and so $$2^{2b-1}=(5^{a''}-3^{a''})(5^{a''}+3^{a''}).$$ Then $$5^{a''}-3^{a''}=2^u$$ and $$5^{a''}+3^{a''}=2^v$$ for integers $$v>u>0$$ such that $$u+v=2b-1$$. But then $$2^u(1+2^{v-u})=2^u+2^v=(5^{a''}-3^{a''})+(5^{a''}+3^{a''})=2\times5^{a''},$$ so $$u=1$$ and $$v=2b-2$$. We see that $$5^{a''}-3^{a''}=2$$ so $$a''=1$$, so $$2^v=5^{a''}+3^{a''}=8$$ and so $$v=3$$. But then $$2b-1=u+v=4$$, a contradiction. This shows that no solution with $$b>a$$ exists.
A similar argument shows that $$b is impossible. This means $$a=b$$ and the equation becomes $$5^c7^c=3^a7^a+4^a7^a=(3^a+4^a)7^a.$$ By unique factorization $$a\leq c$$ and so $$5^c7^{c-a}=3^a+4^a.$$ Again similar arguments show that $$a is impossible. Then $$a=b=c$$ and so we are left with $$3^a+4^a=5^a,$$ which clearly has only the solution $$a=2$$.