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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.

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    $\begingroup$ How about Fermat's Last Theorem? $\endgroup$ – Toby Mak Mar 6 at 12:37
  • $\begingroup$ 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. $\endgroup$ – Aravind Mar 6 at 13:25
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    $\begingroup$ What about it? @TobyMak $\endgroup$ – Aqua Mar 6 at 13:45
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    $\begingroup$ @TobyMak How does Fermat's Last Theorem apply to this? The exponents don't have to be the same. $\endgroup$ – John Douma Mar 6 at 13:46
  • $\begingroup$ See also this similar question. $\endgroup$ – Fabio Lucchini Mar 6 at 14:12
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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 $a=2$.

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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.


[Original answer]

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 $b<a$ 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 $a<c$ is impossible. Then $a=b=c$ and so we are left with $$3^a+4^a=5^a,$$ which clearly has only the solution $a=2$.

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  • $\begingroup$ Thanks a lot! It's really great way of proof! $\endgroup$ – Anatoly Karpov Mar 6 at 16:00

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