4

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.

Etemon
  • 6,437

5 Answers5

3

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.

PM.
  • 5,249
2

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 $64 - 56 = 8$

Will Jagy
  • 139,541
1

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 $$232^2-62^2\times14=2^3\times1=8.$$

Servaes
  • 63,261
  • 7
  • 75
  • 163
1

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 $

lhf
  • 216,483
1

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

$[232^2] - [248^2] + [(62)^2 \times 2]$

$=[(62)^2 \times 2] - [(248)^2 - (232)^2]$

$= [(31)^2 \times 8] - [480 \times 16]$

$= [8 \times (31^2)] - [960 \times 8]$

$= 8 \times [31^2 - 960] = 8.$

user2661923
  • 35,619
  • 3
  • 17
  • 39