Question

The square of an odd positive integer is of the form $\lambda m+1$ for some whole number $m$ and constant $\lambda$. Find $max\left(\lambda \right)$.

Answer 1

According to Euclid division lemma , $a=bq+rwhere0\le r<b$

Here we assume $b=8andr\in [1,7]meansr=1,2,3,.....7$

Then, $a=8q+r$

Case 1 :- $whenr=1,a=8q+1$

squaring both sides,

$a²=(8q+1)²=64²q²+16q+1=8(8q²+2q)+1$

$=8m+1,wherem=8q²+2q$

case 2 :- $whenr=2,a=8q+2$

squaring both sides,

$a²=(8q+2)²=64q²+32q+4\ne 8m+1$ [ means when r is an even number it is not in the form of $8m+1$ ]

Case 3 :- $whenr=3,a=8q+3$

squaring both sides,

$a²=(8q+3)²=64q²+48q+9=8(8q²+6q+1)+1$

$=8m+1,wherem=8q²+6q+1$

You can see that at every odd values of r square of a is in the form of $8m+1$

But at every even Values of r square of a isn't in the form of $8m+1$.

Also we know, $a=8q+1,8q+3,8q+5,9q+7$ are not divisible by $2$ means these all numbers are odd numbers

Hence , it is clear that square of an odd positive is in form of $8m+1$

Hence the answer is 8