the mystery of question six?

let a and b be positive integers. prove (a^2 + b^2)/(a+b) is a square of an integer?

#NumberTheory

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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}`	$\boxed{123}$

Comments

can anyone solve this question? if you can please tell me the solution.

Bardia ebadian - 6 months, 2 weeks ago

Try to study Vieta Jumping method, although there is an elegant geometric proof.

https://brilliant.org/wiki/vieta-root-jumping/

Quest Keeper - 6 months, 2 weeks ago

Well, an interesting statement you have here. Obviously it's not true for all positive integers a and b, but we may find out some cases where the above expression is a perfect square. Let this expressionbe denoted by f:(N×N)->Q, f(a,b)=(a²+b²)/(a+b). First, let a=b. Then f(a,b) simplifies to a. Thus (n²,n²) is an infinite family of solutions for the system. Next we think of the case when a and b are distinct. Now this part is a bit tricky and not easy to visualize as the previous one. However we are going to use a trick to simplify it. Let for some a,b€N, we have (a²+b²)/(a+b)=p/q, where gcd(p,q)=1. Now if we multiply both sides of the equation by pq(product of p and q) we will have (pqa²+pqb²)/(a+b)=p², if we rearrange the lhs a bit by multiplying numerator and denominator by pq we will get, (p²q²a²+p²q²b²)/(pqa+pqb)=p². Now this great, because whenever f(a,b) won't be an integer(forget about square integers) we can reduce it to its simplest form of p/q and always claim that f(pqa,pqb) is a perfect square. To generalize this statement we will say that (m(m+n)(m²+n²),n(m+n)(m²+n²)) will satisfy our statement for all m,n€N

Kushal Dey - 6 months, 1 week 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}`	$\boxed{123}$