## Dhaka Higher Secondary 2010/5

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### Dhaka Higher Secondary 2010/5

Find primes greater than $5$ satisfying the equation: \[11x^{36} - 21x^{10} + 26 x^{2} = 48\]

- Tahmid Hasan
**Posts:**665**Joined:**Thu Dec 09, 2010 5:34 pm**Location:**Khulna,Bangladesh.

### Re: Dhaka Higher Secondary 2010/5

let's take in common

$x^2(11x^{34}-21x^8+26)=48$

so,$x^2 \mid 48$

any value of $x$ as a prime here greater than $5$ is none.

ans.no soln.

one can also do it using trial and error(only putting 1 value will contradict everything)

$x^2(11x^{34}-21x^8+26)=48$

so,$x^2 \mid 48$

any value of $x$ as a prime here greater than $5$ is none.

ans.no soln.

one can also do it using trial and error(only putting 1 value will contradict everything)

বড় ভালবাসি তোমায়,মা

### Re: Dhaka Higher Secondary 2010/5

$11x^36-21x^10+26x^2=48$

or, $x^2(11x^34-21x^8+26)=48$

so, $x^2|48$

hence, $48=2^4 \times 3$

so, we can only have a square

number(4) which divides 48.

But, $x>5$.

So, there is no solution!

or, $x^2(11x^34-21x^8+26)=48$

so, $x^2|48$

hence, $48=2^4 \times 3$

so, we can only have a square

number(4) which divides 48.

**(N.B yet dont know if it is a solution)**But, $x>5$.

So, there is no solution!

**A man is not finished when he's defeated, he's finished when he quits.**