Question

# The square of an odd positive integer is of the form λm+1 for some whole number m and constant λ. Find max(λ).

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Solution

## According to Euclid division lemma , a=bq+rwhere0≤r<b Here we assume b=8andr∈[1,7]meansr=1,2,3,.....7 Then, a=8q+r Case 1 :- when r=1,a=8q+1 squaring both sides, a²=(8q+1)²=64²q²+16q+1=8(8q²+2q)+1 =8m+1,where m=8q²+2q case 2 :- when r=2,a=8q+2 squaring both sides, a²=(8q+2)²=64q²+32q+4≠8m+1 [ means when r is an even number it is not in the form of 8m+1 ] Case 3 :- when r=3,a=8q+3 squaring both sides, a²=(8q+3)²=64q²+48q+9=8(8q²+6q+1)+1 =8m+1,where m=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+1Hence the answer is 8

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