Is it possible to evaluate $232^2-62^2\times14$ by factoring or using identities or...? The expression $232^2-62^2\times14$ can be calculated directly ($53824-3844
\times14=8$). But is it possible to evaluate it for example by factoring or using identities?
Here is what I have tried,
$$(58\times4)^2-62^2\times14=58^2\times16-62^2\times14=(60-2)^2(15+1)-(60+2)^2(15-1)$$
Or
$$58^2\times16-62^2\times14=29^2\times64-31^2\times56=(30-1)^2\times8^2-(30+1)^2(8\times7)$$
But I can't see an elegant way to get $8$ from either of the calculatins.
 A: Using a prime decomposition
$$
\begin{align}
232^2-62^2\times14 &= \\
(2^3 \times 29)^2-(2\times31)^2\times2\times7 &= \\
2^3(8\times29^2-(29+2)^2\times7)&= \\
2^3(29^2-28\times29-28) &= \\
2^3(29-28) &= 8
\end{align}
$$
Line 3 above used $31=29+2$ to facilitate further simplification.
A: this is pretty similar to the answer by lhf.   By solving a Pell equation $u^2 - 14 v^2 = 1$   we find formulas:
$$    x^2 - 14 y^2 = (15x + 56y)^2  - 14 (4x+15y)^2  \; ; \;\; $$
or the reverse direction
$$    x^2 - 14 y^2 = (15x - 56y)^2  - 14 (-4x+15y)^2  \; ; \;\; $$
If $(x,y) = (232, 62), $  the reverse formula says
$$    232^2   - 14 \cdot 62^2  = 8^2 - 14 \cdot 2^2  $$
and this is $64 - 56 = 8$
A: The numbers involved are easy to factor, so take out the common factors to get
$$232^2-62^2\times14=(2^3\times29)^2-(2\times31)^2\times(2\times7)
=2^3\times(2^3\times29^2-7\times31^2).$$
The last factor looks similar to a difference of two squares; note that
\begin{eqnarray*}
2^3\times29^2-2^3\times31^2&=&2^3\times(29^2-31^2)\\
&=&2^3\times(29+31)\times(29-31)\\
&=&-2^4\times60\\
&=&-960,
\end{eqnarray*}
and so
$$2^3\times29^2-7\times31^2=-960+31^2=1.$$
It follows that
$$232^2-62^2\times14=2^3\times1=8.$$
A: A little advanced fun (not that it avoids computation):
$
232^2-62^2\times14=N(232+62\sqrt{14})
$
in $\mathbb Q(\sqrt{14})$.
$
232+62\sqrt{14}=2(15 + 4 \sqrt{14})(4+\sqrt{14})
$
$
15 + 4 \sqrt{14}
$ is a unit, actually a fundamental unit of norm $1$. See this question.
Therefore,
$
232^2-62^2\times14=N(232+62\sqrt{14})=4N(15 + 4 \sqrt{14})N(4+\sqrt{14})=4N(4+\sqrt{14})=4(4^2-14)=8
$
A: Alternative approach:
Need to know that $(31)^2 = 961.$
$(62^2) \times 14$ may be re-written as $[(62^2) \times  4^2] - [(62)^2 \times 2].$
Therefore, the overall computation may be re-expressed as
$[232^2] - [248^2] + [(62)^2 \times 2]$
$=[(62)^2 \times 2] - [(248)^2 - (232)^2]$
$= [(31)^2 \times 8] - [480 \times 16]$
$= [8 \times (31^2)] - [960 \times 8]$
$= 8 \times [31^2 - 960] = 8.$
