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:

m = 6

Step-by-step explanation:

4×6 + 1 = 25 is a perfect square

4×12 + 1 = 49 is a perfect square ( since 12 is the multiple of 6)

Answer 2

Answer:

9 values of m exists

Step-by-step explanation:

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