Question

Let m be a natural number (1<=m<=100). For how many values of m, 4m+1 is a perfect square?

Answer 1

answer : 9

explanation : m be a natural number such that 1 ≤ m ≤ 100.

we have to find the number of values of m for which (4m + 1) is a perfect square

as it is given, 1 ≤ m ≤ 100

so, 5 ≤ 4m + 1 ≤ 401

perfect square between 5 to 401 is ; 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400

now, 4m + 1 is an odd number

so, only odd numbers can be selected.

so, 9, 25, 49, 81, 121, 169, 225, 289, 361

now let's check 4m + 1 = 9 ⇒m = 2

4m + 1 = 25 ⇒m = 6

4m + 1 = 49 ⇒ m = 12

4m + 1 = 81 ⇒ m = 20

4m + 1 = 121 ⇒ m = 30

4m + 1 = 169 ⇒m = 42

4m + 1 = 225 ⇒ m = 56

4m + 1 = 289 ⇒m = 72

4m + 1 = 361 ⇒m = 90

so, m = {2, 6, 12, 20, 30, 42, 56, 72, 90}

there are nine values of m for which (4m + 1) is a perfect square.

Answer 2

$\bf{\underline{\underline{Answer:}}}$

$\bf{∴\:9 \:values\: of\: m\: exists}$

$\bf{\underline{\underline{Step\: by\: step\: explaination:}}}$

let say numbers are

2n and 2n+1 whose square is 4m+1

(2n)^2 = 4n^2 =4m hence can not be equal to 4m+1

(2n+1)^2 = 4n^2 + 4n + 1

= 4(n^2 + n) + 1

comparing with 4m+ 1

${n}^{2} + n = m$

${n}^{2} + n \geqslant 1 \\ n \geqslant 1$

${n}^{2} + n \leqslant 100 \\ n \leqslant 9$

hence 9 such values of m exists

n m 4m+1

1 2 9

2 6 25

3 12 49

4 20 81

5 30 121

6 42 169

7 56 225

8 72 289

9 90 361