For how many positive integers n the expression n²+35

Can you explain that process a little bit closer?
I am confronted with this kind of question type for the very first time.

I had a complete wrong approach. I knew that for a number to be a perfect square it has to have a number of odd integers.
Tried to work around this frame but did not arrive at a solution.

Then I just tried to fit numbers, of course 1 made sense right from the beginning as 1*1 +35 is a perfect square.

I just don't quite get the steps involved in your approach.

You guys rewrite the equation in the form (x+n)*(x-n) = 35 and from there plug in viable numbers?

\(n^2+35 = s^2\)

Or, \(s^2 - n^2 = 35\)

Or, \((s + n)(s - n) = 7*5\)

Now, Solve for n and S , \(s = 6\) & \(n = 1\)

Thus, There are 2 possible values, Answer must be (B)

Hi chrtpmdr

After getting (x+n)*(x-n) = 35
take the possible mutiplication pairs of 35
35 = 35*1 or 7*5
now, equating (x+n)*(x-n) = 35*1
we get x+n = 35 and x-n = 1
solve these two by adding the equations, we get 2x= 36 or x = 18. hence n = 17

similarly, equating (x+n)*(x-n) = 7*5
we get x+n = 7 and x-n = 5
solve these two by adding the equations, we get 2x= 12 or x = 6. hence n = 1

hence there can be 2 positive integers n = 17 or 1 which satisfy the given condition.

We let x^2 be the perfect square. So we have:

n^2 + 35 = x^2

35 = x^2 - n^2

35 = (x - n)(x + n)

We see that we have two integers (x - n) and (x + n) whose product must be 35. Thus, by inspection, we see that we can have a product of 35 by either 5 * 7 or by 1 * 35.

For the first case (5 * 7), we have:

x - n = 5

x = n + 5

and

x + n = 7

x = -n + 7

Setting the two equations equal to each other, we have:

n + 5 = -n + 7

2n = 2

n = 1

For the second case (1 * 35), we have:

x - n = 1

x = n + 1

and

x + n = 35

x = -n + 35

Setting the two equations equal to each other, we have:

n + 1 = -n + 35

2n = 34

n = 17

So, there are two cases: n = 1 and n = 17, such that n^2 + 35 is a perfect square.

Answer: B

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