Isn't the left side bigger?

Number Theory Level 4

$\large { 4 }^{ x }+{ 5 }^{ y }={ 3 }^{ z }$

For non-negative integers $x,y,z$ , let triplets $\left\{ { x }_{ 1 },{ y }_{ 1 },{ z }_{ 1 } \right\} ,\left\{ { x }_{ 2 },{ y }_{ 2 },{ z }_{ 2 } \right\} \ldots \left\{ { x }_{ n },{ y }_{ n },{ z }_{ n } \right\}$ be the solutions to the equation above,

Submit your answer as $\displaystyle \sum _{ i=1 }^{ n }( { { x }_{ i }+{ y }_{ i }+{ z }_{ i } } )$ .

The answer is 4.

1 solution

Akash Deep
Jun 8, 2015

First take the cases for x = 0 , y = 0 both will yield no solution. Now x,y,z are natural no.s so $4^{x}$ ends in 4 or 6 and $5^{y}$ ends in 5 so LHS ends with unit digit 9 or 1 .same applies to Rhs so z is divisible by 4 or gives a remainder 2. Now we take cases. Case1 X = 2j + 1 and z = 4k + 2 so
$5^{y} = (3^{2k + 1} + 2^{2j + 1})(3^{2k + 1} - 2^{2j + 1})$ 5 cannot divide both the terms of the above product so the smaller of the 2 is equal to 1. By this we get k = j = 0 and x = 1, y = 1 and z = 2 Case : 2 X = 2j and y = 4m
In this case we get the equation as;- $5^{y} = (3^{2m} + 2^{2j})(3^{2m} - 2^{2j})$ this gives no solution And so our answer is $\boxed{ 4 }$

Isn't the left side bigger?

The answer is 4.

1 solution

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