Proof by induction: Show that $7|5^{2n}-2^{5n}$ I'm stuck on my inductive step, and can't figure out how to manipulate this algebraically to get the form that I want...
$5^{2k}-2^{5k}=7m,m\in\mathbb{Z}$
\begin{align*}
5^{2k+2}-2^{5k+2}=5^{2k}\cdot 5^{2}-2^{5k}\cdot 2^{2}
\end{align*} 
So i'm not sure what to do now...
 A: I wouldn't do this by induction.  Modular arithmetic is your friend!
$$5^{2n} - 2^{5n} = 25^n - 32^n\equiv 4^n - 4^n \equiv 0 \pmod{7}.$$
(You could turn this into a proof by induction: show that for all $n\in \mathbb{N}$ the remainders upon division by $7$ of $5^{2n}$ and $2^{5n}$ are the same.  But it really wouldn't be that nice.)
A: Maybe another solution for $n=1$ we have $25-32=-7$ this one has the divisor $7$.
Now we assume that $7||5^{2n}-2^{5n}$   for a  $n$ and we will show that this implies $7|5^{2(n+1)}-2^{5(n+1)}$.
The main idea is that $32=25+7$, so we will be able to manipulate it in a multiple of our assumption and add a multiple of $7$.
$$5^{2n+2}-2^{5n+5}=25 \cdot 5^{2n} - 32 \cdot 2^{5n}=25\cdot 5^{2n}- (25+7)\cdot 2^{5n}=
25\cdot (5^{2n}-2^{5n}) -7 \cdot 2^{5n}$$
The first part is divisble through $7$ by our assumption, and the second is a multiple of $7$. Hence it is true.
A: $5^{2n}=(5^2)^n=25^n$  Similarly, $2^{5n}=32^n$
$$\text{So, }5^{2n}-2^{5n}=25^n-32^n$$
Now, 
(1)$(a^n-b^n)(a+b)=a^{n+1}-b^{n+1}+ab(a^{n-1}-b^{n-1})$
$\implies a^{n+1}-b^{n+1}=(a^n-b^n)(a+b)-ab(a^{n-1}-b^{n-1})$
If $(a-b)$ divides $(a^{n-1}-b^{n-1}),(a^n-b^n)$
it will divide $ (a^{n+1}-b^{n+1})$
Now, $(a-b)\mid (a-b)$ and $(a-b)\mid (a^2-b^2)$
(2)$(a-b)(a^n+b^n)=a^{n+1}-b^{n+1}-ab(a^{n-1}-b^{n-1})$
$\implies a^{n+1}-b^{n+1}=(a-b)(a^n+b^n)+ab(a^{n-1}-b^{n-1})$
So, $(a-b)\mid (a^{n+1}-b^{n+1})\iff $ it divides $(a^{n-1}-b^{n-1})$
Now, $(a-b)\mid (a-b)$ and $(a-b)\mid (a^2-b^2)$
Here $a=25,b=32$
A: $$\forall\;a,b\in\Bbb R\;,\;\;a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\ldots+ab^{n-2}+b^{n-1})\implies$$
$$5^{2n}-2^{5n}=25^n-32^n=(25-32)(25^{n-1}+25^{n-2}\cdot32+\ldots+25\cdot32^{n-2}+32^{n-1})$$
and since $\,25-32=-7\,$ we're done and without induction as long as we can assume the first equation above.
A: Set $a=5^2=25$ and $b=2^5=32$. Then 
$$
(a^{n+1}-b^{n+1})-(a^{n}-b^{n}) = (a-1)a^n-(b-1)b^n = 24a^n - 31b^n = 24(a^n-b^n) - 7b^n
$$
and so
$$
a^{n+1}-b^{n+1} = 25(a^{n}-b^{n}) - 7b^n
$$
Thus, if $7$ divides $a^{n}-b^{n}$ then $7$ divides $a^{n+1}-b^{n+1}$.
A: $\begin{eqnarray}{\bf Hint}&&\rm n\mid\ \  \color{#C00}{a^k\!-\!b^k}\:\  and\:\  n\mid \color{#0A0}{a-b} \\ \Rightarrow\: &&\rm n\mid (\color{#C00}{a^k\!-b^k})\,a\,+\,b^k(\color{#0A0}{a-b})\, =\, a^{k+1}\!-b^{k+1}\end{eqnarray}$
Note $\ $ The above proof is essentially a special case of that for the Congruence Product Rule. Below are proofs of this product rule proof expressed in both divisibility and congruence form, $ $ using the standard notation: $\rm\  \ a\mid b \ :=\ a\,$ divides $\rm\, b,\ $ and $\rm\ \, a\equiv b\ \ (mod\ n) \iff n\mid a-b$
$\begin{eqnarray}
\rm {\bf Lemma}\ \   &\rm n\ \ |&\rm\ \, X-x\quad\ and &&\rm n\ |\: Y-y \ \Rightarrow\ n\:|\!\!&&\rm XY - \: xy\\ \\
\rm {\bf Proof}\ \ \ \ \ &\rm n\ \ |&\rm (X-\color{brown}x)\:\color{brown}Y\ \ \ +   &&\rm\, \color{brown}x\ (\color{brown}Y-y)\ \ \ = &&\rm XY - \: xy \\
\\
\rm {\bf Lemma}\ \ &         &\rm\ \, X\equiv x\quad\ \ and &&\rm\quad\ \ Y\equiv y \ \ \ \ \Rightarrow\ &&\rm XY\equiv xy\\ \\
\rm {\bf Proof}\ \ \ \ \ &0\equiv& \rm (X-\color{brown}x)\:\color{brown}Y\ \ \ + &&\rm\, \color{#C00}x\ (\color{brown}Y-y)\ \ \ \equiv &&\rm XY - \: xy \\
\end{eqnarray}$ 
