# A note on Diophantine problems

In solving Diophantine problems, one must guard against falling into certain formal traps. The identity{} $(a+b)^{2}=a^{2}+2ab+b^{2}$

might lead one to think that, because all numbers of the form $a^{2}+2ab+b^{}{2}$ are squares, any number of say, the form $a^{2}+3ab+b^{2}$ could never be a perfect square because it is not an algebraic square. But, numbers are more versatile than that. Try $a=7, b=3$ in the expression $a^{2}+3ab+b^{2}$.

It was once proposed as a theorem that “the product of four consecutive terms of an AP of integers plus the fourth power of the common difference is always a perfect square but never a perfect fourth power.” If a is the first term of the four and b the common difference, we have $a(a+b)(a+2b)(a+3b)+b^{4}$ which works out to be the same thing as $(a^{2}+3ab+b^{2})^{2}$.

The truth of the first assertion of the theorem is now evident, but we have already given a counter example showing the second assertion to be incorrect.

Ref: Excursions in Number Theory, C. Stanley Ogilvy and John T. Anderson,

http://www.amazon.in/Excursions-Number-Theory-Dover-Mathematics/dp/0486257789/ref=sr_1_4?s=books&ie=UTF8&qid=1491402900&sr=1-4&keywords=excursion+in+mathematics

Nalin Pithwa.

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