Olympiad Corner

5) If $a, b, c, d, e$ are natural numbers, then prove that $a^2 + b^3 + c^4 + d^5 = e^6$ has infinite many solutions.

#NumberTheory

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

a solution is (2,3,1,2,2). ssince there exists a solution, we multiply both sides by $2^{60}$ which gives us $(2^{30}a)^2+(2^{20}b)^3+(2^{15}c)^4+(2^{12}d)^5=(2^{10}e)^6$ we can repeat this process infinitly giving us infinite solutions. hence solved.

Aareyan Manzoor - 5 years, 6 months ago

@Aareyan Manzoor can you give the primitive solution!

Sualeh Asif - 5 years, 6 months ago

i tried but couldnt find one. let me write a code, i will have a pair in a while...

Aareyan Manzoor - 5 years, 6 months ago

(2,3,1,2,2) @Sualeh Asif

Aareyan Manzoor - 5 years, 6 months ago

Nice

Abdur Rehman Zahid - 5 years, 6 months ago

$22^2+1^3+1^4+3^5=3^6$ Hence (22,1,1,3,3) is another solution

Abdur Rehman Zahid - 5 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$