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Daniel Wang
- 7 years, 9 months ago

$24^4-1=(24^2+1)(24^2-1)=(24^2+1)(24+1)(24-1)$ . We would show that $24^2+1=577$ is a prime.

Note that it is not divisible by 3, 5, 7, 11, 13, 17, 19, or 23. Since $\sqrt{577}<29$ , we know now that it is a prime. Hence the answer is $\boxed{577}$ .

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We know that 577 is the largest possible prime since the other 2 factors, 25 and 23 are both less than 577.

Russell FEW
- 7 years, 9 months ago

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But what if $577$ wasn't a prime??? We couldn't be sure whether $24^2 + 1$ is a prime or composite... If $577$ wasn't a prime, we would need the other prime factors to continue... Prime factorization should make the solution more clear...

Abrar Nihar
- 7 years, 9 months ago

@ Calvin: Agreed. I think that in addition to that I should have done all the trial divisions by 3, 5, 7, 11, 13, 17, 19 and 23 to 577.

Russell FEW
- 7 years, 9 months ago

From $a^{2}$ - $b^{2}$ =( $a + b$ )( $a - b$ )

We know

$24^{4}$ - $1^{4}$ = ( $24^{2}$ + $1^{2}$ ) ( $24^{2}$ - $1^{2}$ ) = ( $24^{2}$ + $1^{2}$ ) ( $24 + 1$ ) ( $24 - 1$ ) = $577 \times 25 \times 23$

Hence, 577, 25 and 23 are factors of $24^{4}$ - $1^{4}$ ,and only 577 and 23 are prime factors, thus the largest prime factor is 577

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$\begin{aligned} 24^4 - 1 & = \left(24^2 + 1\right)\left(24^2 - 1\right) \\ & = \left(577\right)\left(24 + 1\right)\left(24 - 1\right) \\ & = \left(577\right)\left(5^2\right)\left(23\right) \end{aligned}$

Here is the equation, $\boxed{577}$ is the largest prime, thus it is our answer.

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You should verify that 577 is indeed prime.

$=24^{4}-1$

$=(24^{2}+1)(24^{2}-1)$

$=(24^{2}+1)(24+1)(24-1)$

$=(577).(25).(23)$

Since 577 is prime, so the largest prime factor is 577

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Se Repararmos Isto é um Quadrado da Soma Pela Diferença :

(24^2+1) \times (24^2-1) (24^2+1) \times (24 +1) \times (24-1) (577) \times (5^2) \times (23)

Assim Temos o Maior Primo = 577

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let 24=x

x^4-1=(x^2+1)(x^2-1) =(x^2+1)(x+1)(x-1) =(24^2+1)(24+1)(24-1) =(577)(25)(23)

because 577 is prime number,so,the largest prime factor of 24^4-1 is 577

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24^4 - 1 = 331776 - 1 = 331775.

the prime factorization of 331775 = 5^2 . 23 . 577.

then the largest prime factor of 24^4 - 1 = 577

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wonderfull solution: http://www.mathwarehouse.com/arithmetic/numbers/prime-number/prime-factorization-calculator.php

Andrei Popescu
- 7 years, 8 months ago

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24^4-1= (24^2+1)(24^2-1). The greater prime factor is 24^2+1= 577.

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Since 24^4 = 576^2, so (576^2)-1 = 575*577. Thus 577 is the largest prime factor of (24^4)-1.

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24^4 - 1

```
= (24^2 - 1) * (24^2 + 1)
= (24 - 1) * (24 + 1) * (24^2 + 1)
= 23 * 5^2 * 577
```

Where the grates prime factor is 577.

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24^4-1=331776-1=331775=5×66355=5×5×13271=5×5×23×577

or, 24^4-1=24^4-1^4=(24^2+1^2)×(24^2-1^2)=(24^2+1^2)×(24-1)×(24+1)=577×23×25=577×23×5×5

so, 577 is the largest prime factor of 24^4-1

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again: http://www.mathwarehouse.com/arithmetic/numbers/prime-number/prime-factorization-calculator.php

Andrei Popescu
- 7 years, 8 months ago

$24^4-1=5^2 \cdot 577 \cdot 23$

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×

Problem Loading...

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$~~~~~~~\Large{24^4-1}$

$\Large{=(24^2)^2-1^2}$

$\Large{= (24^2+1)(24^2-1)}$

$\Large{=(24^2+1)(24+1)(24-1)}$

$\Large{=577 \times 25 \times 23}$

$\Large{=5^2 \times 23 \times 577}$

$~$

$\Large{\textrm{We can see that}⌊\sqrt{577}⌋ = 24 }$

$~$

$\Large{\textrm{Diving 577 by every prime below 24...}}$

$\Large{\textrm{We can find that 577 is a prime}}$

$~$

$\Large{\textrm{Hence, the largest prime is}}$ $\LARGE{\fbox{577}}$