Double Pythagoras!

Number Theory Level 4

$1458000=a^{2}+b^{2}=c^{2}+d^{2}$ where $a,b,c$ and $d$ are positive integers. Find $a+b+c+d$ .

Note: Please refrain from using Wolfram Alpha.

The answer is 3024.

1 solution

Ricardo Moritz Cavalcanti
Aug 21, 2020

We first note that

$1458000 = 2^4 3^6 5^3 = 5(2^2 3^3 5)^2 = 125(2^2 3^3)^2$ .

Now, let $a=2^2 3^3 5 A$ and $b=2^2 3^3 5 B$ ; then

$a^2+b^2=1458000 \Rightarrow (2^2 3^3 5)^2(A^2+B^2)=5(2^2 3^3 5)^2 \Rightarrow A^2+B^2 = 5$ ,

a possible solution of which is $A=1$ , $B=2$ , so that $a=2^2 3^3 5\cdot 1=540$ and $b=2^2 3^3 5\cdot 2=1080$ .

In a similar way, let $c=2^2 3^3 C$ and $d=2^2 3^3 D$ ; then

$c^2+d^2=1458000 \Rightarrow (2^2 3^3)^2(C^2+D^2)=125(2^2 3^3)^2 \Rightarrow C^2+D^2 = 125$ ,

a possible solution of which is $C=2$ , $D=11$ , so that $c=2^2 3^3\cdot 2 = 216$ and $d=2^2 3^3\cdot 11 = 1188$ .

Finally, $a+b+c+d=540+1080+216+1188=3024. \,\square$

Why do you think only $a = 1, b = 2, c = 2, d = 11$ ? Reversing the values also fit the question. Like $a = 2, b = 1, c = 11, d = 2$ .

. . - 3 months, 3 weeks ago

I don't --- note that I wrote "a possible solution" ---, but I didn't bother to write the other solutions because permutations of {a,b} and {c,d} don't change the value of a+b+c+d.

Ricardo Moritz Cavalcanti - 3 months, 3 weeks ago

Double Pythagoras!

The answer is 3024.

1 solution

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