Observe 1

Number Theory Level 2

$a$ , $b$ , $c$ , and $d$ are distinct non-negative integers such that $a^2+b^2=c^2+d^2=e^2$ , where $e$ is a non-negative integer.

For the minimum value of $e^2$ , what is the value of $a+b+c+d+e$ ?

17 28 10 34

2 solutions

Mr. India
Mar 2, 2019

$25=5^2=5^2+0^2=3^2+4^2$

$a+b+c+d+e=0+5+3+4+5=17$

$25$ is minimum as it is the smallest perfect square which can be written as sum of 2 other perfect squares and also fulfilling the assumption

I have amended the problem wording for you. We have to mention $a$ , $b$ , $c$ , and $d$ are distinct. If not there will be infinitely many solutions.

Chew-Seong Cheong - 2 years, 3 months ago

Sir, how would there be infinite solutions?? And how did you amended problem? I had mentioned the assumption for this reason only.

Mr. India - 2 years, 3 months ago

$0^2+0^2=0^2+0^2 = 0^2$ , $0^2+1^2=0^2+1^2=1^2$ ,,, I noted you have mentioned "only two can be equal". But it is clearer to just say $a$ , $b$ , $c$ , and $d$ are distinct. You don't need to bring in $e$ .

Chew-Seong Cheong - 2 years, 3 months ago

@Chew-Seong Cheong – Ok,thank you sir. But how could you edit the question?

Mr. India - 2 years, 3 months ago

@Mr. India – I am a moderator. If you are active in Brilliant you may be appointed as a moderator. I have added "where $e$ is a non-negative integer."

Chew-Seong Cheong - 2 years, 3 months ago

@Chew-Seong Cheong – OK sir thank you!

Mr. India - 2 years, 3 months ago

Chew-Seong Cheong
Mar 2, 2019

Considering $a^2 + b^2 = c^2 + d^2 = e^2$ , to have the smallest $e^2$ , it is reasonable for us to assume one of the four distinct non-negative integers be the smallest 0. Let $d=0$ . Then $a^2 + b^2 = c^2 = e^2$ , and $c=e$ . It is obvious that $(a, b, c)$ is the smallest set of Pythagorean triples $(3,4,5)$ . Therefore, $a+b+c+d+e =$ $3+4+5+0+5 = \boxed{17}$ .

Observe 1

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