If $ (1+3+5+\cdots +p)+ (1+3+5+\cdots +q)= (1+3+5+\cdots +r)$, Find the minimum value of $p+q+r$ if $p>6$
Let $p=2a-1$, $q=2b-1$, $r=2c-1$
$$(1+3+5+\cdots +p)+ (1+3+5+\cdots +q)= (1+3+5+\cdots +r)$$ $$\implies a^2+b^2=c^2$$ I have to minimize $f(a,b,c)=2a+2b+2c-3$ subject to $a^2+b^2=c^2$
On using Lagrange multipliers, $$\frac{\partial f}{\partial x}=k\frac{\partial g}{\partial x}$$ $$\implies 2=2ka$$ $$k=\frac{1}{a}$$ Similarly, $k=\frac{1}{b},k=\frac{1}{c}$. So $a=b=c$. But $a^2+b^2=c^2$. Hence $a=b=c=0$. This is not possible.
p=7 -> S(p=7) = 16
q=5 -> S(q=5) = 9
r=9 -> S(r=9) = 25
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p=23 -> S(p=23) = 144
q=17 -> S(q=17) = 81
r=29 -> S(r=29) = 225
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p=29 -> S(p=29) = 225
q=15 -> S(q=15) = 64
r=33 -> S(r=33) = 289