Let the two consecutive positive integers be xx && x+1.x+1.
x^2+(x+1)^2=365x2+(x+1)2=365
We know that (a+b)^2=a^2+2ab+b^2.(a+b)2=a2+2ab+b2.So the equation becomes
=>x^2+(x^2+2*x*1+1^2)=365⇒x2+(x2+2⋅x⋅1+12)=365
=>x^2+x^2+2*x*1+1^2=365⇒x2+x2+2⋅x⋅1+12=365
=>x^2+x^2+2x+1=365⇒x2+x2+2x+1=365
=>2x^2+2x+1=365⇒2x2+2x+1=365
=>2x^2+2x+1color(red)(-365)=365color(red)(-365)⇒2x2+2x+1−365=365−365
=>2x^2+2x-364=0⇒2x2+2x−364=0
Apply Quadratic Formula, that is x=(-b+-sqrt(b^2-4ac))/(2a).x=−b±√b2−4ac2a. Here
a=2, b=2, c=-364a=2,b=2,c=−364
=>x=(-2+-sqrt(2^2-(4*2*(-364))))/(2*2)⇒x=−2±√22−(4⋅2⋅(−364))2⋅2
=>x=(-2+-sqrt(4-(8*(-364))))/(4)⇒x=−2±√4−(8⋅(−364))4
=>x=(-2+-sqrt(4-(-2912)))/4⇒x=−2±√4−(−2912)4
=>x=(-2+-sqrt(4+2912))/4⇒x=−2±√4+29124
=>x=(-2+-sqrt(2916))/4⇒x=−2±√29164
=>x=(-2+54)/4⇒x=−2+544 oror x=(-2-54)/4x=−2−544
=>x=52/4⇒x=524 oror x=(-56)/4x=−564
=>x=13⇒x=13 oror x=-14x=−14
xx won't be negative because the value must be a positive integer.
So, x=13x=13
Then x+1=14x+1=14
So the positive consecutive integers whose sum of squares is 365 are 1313 && 14.14.
=>13^2+14^2=365⇒132+142=365
=>169+196=365⇒169+196=365
=>365=365⇒365=365