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Class 10 > Learn Mathematics > If m and n are odd positive integers, then m2 + n2...
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If m and n are odd positive integers, then m2 + n2 is even, but not divisible by 4. Justify.

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Let the two odd positive numbers be x = 2k + 1 a nd y = 2p + 1 Hence x2 + y2 = (2k + 1)2 + (2p + 1)2 = 4k2 + 4k + 1 + 4p2 + 4p + 1 = 4k2 + 4p2 + 4k + 4p + 2 = 4(k2 + p2 + k + p) + 2 Clearly notice that the sum of square is even the number...
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Let the two odd positive numbers be x = 2k + 1 a nd y = 2p + 1 Hence x2 + y2 = (2k + 1)2 + (2p + 1)2 = 4k2 + 4k + 1 + 4p2 + 4p + 1 = 4k2 + 4p2 + 4k + 4p + 2 = 4(k2 + p2 + k + p) + 2 Clearly notice that the sum of square is even the number is not divisible by 4 Hence if x and y are odd positive integers, then x2 + y2 is even but not divisible by 4 cross check let m= 3 and n=5 3^2+5^2=34 34 is not divisible by 4 hence prove read less
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Since m is odd m*m is also odd. This imply m*m+n*n is of the form 4*k+2 (Since m*m is of the form 4*k+1).
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Since m is odd positive integer then we can write m=2a+1 and n=2b+1, such that a,b >0 m2+n2=(2a+1)^2+(2b+1)^2 = 4a^2+1+4a+4b^2+1+4b = 4a^2+4b^2+4a+4b+2 = 2(2a^2+2b^2+2a+2b+1) Above eqn is divisible by 2 but not divisible by 4.
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Let m=2a+1, n=2b+1 be two odd positive integers Now m^2+n^2=(2a+1)^2+(2b+1)^2 =4a^2+1+4a+4b^2+1+4b =4a^2+4b^2+4a+4b+2 =2(2a^2+2b^2+2a+2b+1) Therefore m^2+n^2 is divisible by 2 but not with 4 Example let m=1, n=5 m^2+n^2=(1×1)+(5×5)=1+25=26 26 is divisible by 2 but not with 4
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M=1,n=3 then m2+n2=one2+three2=10,10 is not divisible by 4
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